How many seconds are there in a meter?

[Frylock]

Is there an answer to the question in the title of my post? If time is a dimension in the same sense as the spatial dimensions are dimensions, then I might think there should be a sensible way to convert meters into seconds and vice versa.

Well, is there?

-FrL-

[ParentalAdvisory]

I can think of measuring the amount of time an object takes to get from point A to point B in a meter. I think it most cases, movement is required to make that measurement. Otherwise, the meter (or any measure of length) is just here and now. How could you get time out of it?

[Tangent]

As I understand it, time is not exactly the same as the spatial dimensions, so I don't think there really is an answer to your question.

The best that I can do is tell you that the meter is currently defined as the distance light travels in 1/299,792,458 of a second.
(links to Wikipedia entry on the meter)

[susan]

I was taught by friends at MIT that the answer to any question like this is "11 Bonner Units."

[tomndebb]

I suppose that one could identify one second as equal to 299,792,458 meters, (the number of meters that light travels in 1 second in a vacuum).

[Pasta]

c.

That is, there are 299,792,458 meters in one second.

In relativity, time and space are on the same footing, units-wise. The meters/seconds distinction is a Newtonian construct.

It takes massless particles a time of 1 unit to travel a distance of 1 unit. Since the time unit of 1 second is equal to a distance of 299,792,458 meters, light travels at 299,792,458 meters/second = 1 distance unit / 1 time unit.

[DanBlather]

Hown many inches of hieght are there in a inch of width?

[Orbifold]

I've got to agree with Tangent (and disagree with Pasta). In special relativity time is one dimension and space is the other three, but that doesn't exactly mean they're interchangeable.

Let me 'splain.

Considering just space for a moment, 1 metre "up" is the same as 100 centimetres "to the left"; specifically, if I have two objects, one of which is 1 metre "up" from the other, I can rotate the whole system rigidly in such a way that I end up with two objects, one of which is 100 centimetres "to the left" of the other.

Now consider space-time. I can conceive of two events in space-time, one of which is 1 second after the other but in the same location with respect to an observer. I can also conceive of two events which take place at the same time with respect to an observer, but one of which takes place 1 metre to the left of the other. But there's no such thing as a "rigid" motion of the system which takes the first configuration of events to the other. You just can't rotate time to point in the same direction as space, any more than you can accelerate an object past the speed of light.

So there's no real need to be able to convert seconds into meters, even though they exist as two dimensions of same space-time. You can just arbitrarily declare that 1 second equals c meters, where c is the speed of light, and that simplifies certain calculations, but that just a convenience; there's nothing physically necessary about that conversion in the same way that the rate of 1 meter to 100 centimeters is necessary.

That said, general relativity is a whole other kettle of fish, and there may be a natural answer to your question there. But I don't claim to understand general relativity beyond the whole "bowling balls on a trampoline" thing.

[Frylock]

Thanks for the replies so far, and thanks especially to [b]Orbifold[/i].

-FrL-

[Pasta]

Originally Posted by Orbifold

I've got to (disagree with Pasta).

Yes, space and time are not the same. That's why I took care to say "units-wise". But, while it's true that there's no spatial rotation that takes distance into time, there are Lorentz "rotations" that do. Space and time do indeed mix, you just need a more general rotation to do it. And, if you express them in the same units, they differ only by a minus sign.

Or in other words: the question in the OP has an answer only in the context of space-time, and in the context of space-time space and time are on the same footing when it comes to their units. They differ only by a minus sign in the equations (in special relativity that is.)

[Frylock]

Special thanks extended also to Pasta.

Now I don't know who to believe, though.

-FrL-

[David Simmons]

As Orbifold said, just because time and locations form a four dimensional coordinate system it doesn't mean that there is an equivalence between time and location.

For example, we agree to meet in the building at the corner of 1st Street and 1st Avenue at the top of the stairway on the 3rd floor. We have now given the location of a point in three dimensional space where we will meet. There is, however, one vital dimension missing that is required in order for us to meet. There is the little matter of when, or at what time, we are going to meet. So, in order to specify the location of a point in a universe where there is movement you really need to know the time at which an object was at a particular point in three dimensional space.

It turns out that because of the finite speed of light, when you specify the three spacial locations and the time for a moving object as seen by one observer, all four of them change when you refer then to a different observer. That is time and space are inextricably interwolven in a single coordinate system.

[Orbifold]

Originally Posted by Pasta

Yes, space and time are not the same. That's why I took care to say "units-wise". But, while it's true that there's no spatial rotation that takes distance into time, there are Lorentz "rotations" that do.

No, I have to disagree. Unless you're using the term "Lorentz rotation" in a way I'm not expecting.

A rigid motion of Minkowski space...space-time as viewed by special relativity, in other words...has to preserve the "spacetime distance". That means it has to preserve the set of light cones, because light cones are just the sets of points where the spacetime distance between any two points is zero. And light cones have a definite "direction", in that the time axis is always on a different side of the light cone than the three spatial axes.

In short, any homeomorphism of Minkowski space which swaps the time axis with one of the space axes would not preserve the set of light cones and hence can't possibly be an isometry with respect to the space-time metric.

[Pasta]

More detail...

In a Newtonian world, one can rotate width (x) and depth (y) into a different arrangement like so:

x' = x*cos(q)+y*sin(q)
y' = -x*sin(q)+y*cos(q)

where q is the angle of rotation. Note that you have two quantities (x and y) that each get multiplied by unitless numbers to create the new quantities. Everything is on the same footing: x, y, x', and y'. Time has no place here since it's Newtonian.

In special relativity, you could also do this rotation. But, you can also rotate space and time into one another. Taking x and t now:

ct' = ct*g+x*gb
x' = -ct*gb+x*g

Note the similarity between this and the purely spatial rotation. g and b are unitless numbers describing the extent of the rotation. The conversion constant c is only there because humans decided to use different units for space and time. In the spatial example, if I had chosen to measure x in inches and y in centimeters, and I had decided to pretend that these were fundamentally different things, I would have needed a conversion constant k=(2.54 cm/in) in the expression:

kx' = kx*cos(q)+y*sin(q)
y' = -kx*sin(q)+y*cos(q)

But we know that trying to add cm and inches requires an implicit conversion, so we don't write it explicitly. The same holds for the space-time situation. If we had just expressed time in meters in the first place, the expression would look clean:

t' = t*g+x*gb
x' = -t*gb+x*g

Replying to items on preview:

Originally Posted by Orbifold

In short, any homeomorphism of Minkowski space which swaps the time axis with one of the space axes <snip>

Indeed, that is not a valid Lorentz transformation, so the light cones are busted. I'm not saying time and space are the same. But they don't need to be interchangeable with respect to the physicical laws to require they have the same units.

Do you not agree that time and space show up in special relativity on equal unit-footing? (Not that they're interchangeable -- they're not!) If you do not, then perhaps we're caught on semantics, because I know we both know the physics.

[David Simmons]

Originally Posted by Orbifold

A rigid motion of Minkowski space...space-time as viewed by special relativity, in other words...has to preserve the "spacetime distance".

Right. In space coordinates the distance between two points is invariant regardless of your coordinate system. That invariant in rectangular coordinates is x² + y² + z². In space time it is t² - x² - y² - z²

[Pasta]

Originally Posted by David Simmons

In space time it is t² - x² - y² - z²

Perfect! And that expression makes sense only if t and x have the same units! Which brings us all the way back: 1 second = 299,792,458 meters.

[Orbifold]

Originally Posted by Pasta

Indeed, that is not a valid Lorentz transformation, so the light cones are busted. I'm not saying time and space are the same. But they don't need to be interchangeable with respect to the physicical laws to require they have the same units.

Do you not agree that time and space show up in special relativity on equal unit-footing? (Not that they're interchangeable -- they're not!) If you do not, then perhaps we're caught on semantics, because I know we both know the physics.

I don't think it's entirely semantics. You say that the physical laws require that they have the same units; that's the part I disagree with. The physical laws don't even require that we have a way to convert between the units. It's convenient to have a conversion between the units, because then the formulas become cleaner, as you say. For example, the bilinear form which defines the metric becomes -t²+x²+y²+z².

But that's just a convenience. It's not necessary. There's nothing preventing us from using the bilinear form -t²+(x/c)²+(y/c)²+(z/c)² instead, where c=299,792,458, and leaving x, y, and z in metres while t remains in seconds. This gives us exactly the same results. It's just not as pretty, because we keep having to insert 299,792,458 into equations all over the place, but it works, and we never have to convert between meters and seconds.

We could do the same thing in just space, and try to define distance in the plane using the formula x²+(y*2.54)² where x is in centimetres and y is in inches. That would be a perfectally valid way to define Euclidean geometry in the plane. But in that case we have an isometry which takes the x-axis to the y-axis, and that provides a physical reason to say that 1 inch is 2.54 centimetres. There's no isometry in Minkowski space which takes a space-like vector to a time-like one, however, so there's no physical reason to say that 1 second is 299,792,458 metres.

It's just really, really handy to do so for computational reasons.

[FRDE]

There is a straightforward answer to this

[Pasta]

I grant your point. The OP's question can only be addressed if one is willing to place an artificial constraint on the form of the bilinear. The answer, then, is as artificial as the constraint. That is, the answer is not a physical law but a metric-dependent statement. A relevant statement, given the ubiquitousness and Occam-pleasingness of the "+/-1" metric, but yes, a non-fundamental one.

[FRDE]

Drat, hit post in error

http://www.boatersdream.com/libartic...TOKEN=18047320

NAUTICAL MILE - One minute of latitude; approximately 6076 feet - about 1/8 longer than the statute mile of 5280 feet.

So a Second of latitude is 6076 / 60 = 101.27 feet

So there are about 30 metres in a Second

[Malacandra]

Time and space are functionally identical and can be meaningfully described in the same units; and of this I have found the most marvellous proof -

Wait one, someone's just come up the road from Porlock and I can't post while he's banging at the door.

[msmith537]

What is the sound of one hand clapping...

SPOILER:

...you across the face for asking such a question!

?

[RealityChuck]

You have to understand what is meant by "dimension."

Simply put, a dimension is a number required in order to locate a point, given a starting point.

Thus, in a one-dimensional space, you only need one number to locate the point. In two-dimensional space, you need two numbers. In three-dimensional space, three, etc.

But these numbers need not be linear measurements. For instance, in a two-dimensional space, you can locate the point given an angle and distance. Thus the point at 2, 2 on a plane can also be located with an angle of 45 degrees and a distance of the square root of eight (assuming my math is correct -- but you get the idea).

Now, lets assume in our two dimensional world that you are trying to hit a moving target. Point 2,2 is fine if the object is stationary, but to hit the moving target, you need to include the time where the target is at point 2,2. Time thus becomes a dimension.

And we use time as a dimension all the time. "It's a one-hour drive," for instance, is one way of locating a position. (Note that we often leave out dimensions for convenience sake. You would say, "It's five miles to the gas station," and ignore the depth and breadth, since they don't matter and a road can be considered to be a line, even though, technically, it's three dimensions, not one.)

[Bosda Di'Chi of Tricor]

Originally Posted by Frylock

How many seconds are there in a meter?

The # of seconds in a meter varies, based on the ammount of energy expended in traversing the meter.

[Frylock]

Originally Posted by RealityChuck

And we use time as a dimension all the time. "It's a one-hour drive," for instance, is one way of locating a position. (Note that we often leave out dimensions for convenience sake. You would say, "It's five miles to the gas station," and ignore the depth and breadth, since they don't matter and a road can be considered to be a line, even though, technically, it's three dimensions, not one.)

Do I understand correctly that you're point here is to offer a case in which we use a "time measure" in order to indicate a "space distance?" So for example, in this case, the answer to my OP would be something like "In this case there are five miles in an hour"?

-FrL-

[Frylock]

Originally Posted by Pasta

I grant your point. The OP's question can only be addressed if one is willing to place an artificial constraint on the form of the bilinear. The answer, then, is as artificial as the constraint. That is, the answer is not a physical law but a metric-dependent statement. A relevant statement, given the ubiquitousness and Occam-pleasingness of the "+/-1" metric, but yes, a non-fundamental one.

Thanks for all the discussion.

It looks like you guys are zeroing in on saying the answer to the question "How many meters in a second" could be anything, depending on what arbitrary equivalence you've agreed to use beforehand.

If that's right, then just to make this more clear to me, could you give me an example as to how I could, so to speak, "make" one second equal one meter, and then an example as to how I could, so to speak, "make" one second equal two meters?

-FrL-

[Sapo]

Originally Posted by FRDE

There is a straightforward answer to this

this was the best answer by far. Until you clarified it was an error.

[Pasta]

Originally Posted by Frylock

If that's right, then just to make this more clear to me, could you give me an example as to how I could, so to speak, "make" one second equal one meter, and then an example as to how I could, so to speak, "make" one second equal two meters?

The numerical answers given above all set 1 time unit equal to the distance a massless particle travels in that time unit. That happens to be the velocity c that shows up in the equations as written typically. But, if your equations had a bunch of twos all over the place, then you might have set c'=c/2, and you might then have decided it sensible to equate 1 time unit with half the distance a massless particle travels in that time unit. 2 and c' always show up multiplied together, so it's an obvious simplification to choose a unit equivalence that casts away the product 2c' rather that one which casts away only the c' (leaving twos everywhere), but Orbifold points out that there's no fundamental reason to choose one over the other. Nature doesn't care how we choose to relate the meter and the second.

So, if you want to equate 1 s = 1 m, the bilinear (and everything else) needs to reflect the choice:

1 s = 1 m
ds² = (299,792,458*dt)²-dx²-dy²-dz²

1 s = 2 m
ds² = (149,896,229*dt)²-dx²-dy²-dz²

1 s = 299,792,458 m
ds² = dt²-dx²-dy²-dz²

[Will Repair]

One second is equal to the number of meters the universe has expanded in that second.

[4.66]

(yet another misinterpretation of your question)

Minutes and seconds are just a way of dividing units into smaller units based on the sexagecimal system (base 60). An hour is divided in to 60 minutes, which is subdivided into 60 seconds. There are also 3600 seconds in a degree of latitude. So there are 3600 seconds in a meter in exactly the same way there are 1000 millifurlongs in a furlong.

[Bryan Ekers]

I dunno, but there are usually two in a duel.

[Ring]

Regardless of the above, all relativists set c equal to 1. Imagine a graph with the vertical axis as time and calibrated in years, and the horizontal axis as distance with units of lightyears.

The Plot of something moving at c would be a line at 45 degrees with a slope of 1. So 1 year would equal 1 lightyear. From this it is very simple to figure out how many meters are in a second.

But this is all just because relativists are lazy.

[Ring]

Actually, in the above I should have said in the world where c =1,1 year can equal 1 lightyear.

But this is all just because relativists are lazy. I.e.

E² = m²c⁴ + p²c²

Becomes E² = m² +p²

or if and only if p =0

E = mc²

Becomes

E =m

[jackdavinci]

This is probably entirely incorrect for reasons that I can't being to comprehend. But it sounds really nice so I'll propose it anyway and legitimize it under the terms of my artistic license.

The smallest measurable length (Planck length) is 1.6162412 x 10^-35 meters and the smallest measurable time (Planck time) is 5.3912140 x 10^-44 seconds. Assuming they are equal, it works out to about

299791698 meters = 1 second
(which is interestingly similar to the speed of light at 299792458 m/s)

or

3.33564 x 10^-9 seconds = 1 meter

and in fact, if you go back to how Planck length and time are derived, if you were to divide them the terms would indeed cancel to C, the speed of light. any apparent disparity is due to rounding errors.

[Contrapuntal]

Originally Posted by tomndebb

I suppose that one could identify one second as equal to 299,792,458 meters, (the number of meters that light travels in 1 second in a vacuum).

Not to go off on a Tangent, or anything.

[Tangent]

Originally Posted by Contrapuntal

Not to go off on a Tangent, or anything.

[Keeve]

Originally Posted by H. G. Wells

'You must follow me carefully. I shall have to controvert one or two ideas that are almost universally accepted. The geometry, for instance, they taught you at school is founded on a misconception.'

'Is not that rather a large thing to expect us to begin upon?' said Filby, an argumentative person with red hair.

'I do not mean to ask you to accept anything without reasonable ground for it. You will soon admit as much as I need from you. You know of course that a mathematical line, a line of thickness *nil*, has no real existence. They taught you that? Neither has a mathematical plane. These things are mere abstractions.'

'That is all right,' said the Psychologist.

'Nor, having only length, breadth, and thickness, can a cube have a real existence.'

'There I object,' said Filby. 'Of course a solid body may exist. All real things--'

'So most people think. But wait a moment. Can an *instantaneous* cube exist?'

'Don't follow you,' said Filby.

'Can a cube that does not last for any time at all, have a real existence?'

Filby became pensive. 'Clearly,' the Time Traveller proceeded, 'any real body must have extension in *four* directions: it must have Length, Breadth, Thickness, and -- Duration. But through a natural infirmity of the flesh, which I will explain to you in a moment, we incline to overlook this fact. There are really four dimensions, three which we call the three planes of Space, and a fourth, Time. There is, however, a tendency to draw an unreal distinction between the former three dimensions and the latter, because it happens that our consciousness moves intermittently in one direction along the latter from the beginning to the end of our lives.'

from Chapter 1 of The Time Machine. U.S. Copyright has expired; full text available online at Project Gutenberg

[Frylock]

Originally Posted by Keeve

from Chapter 1 of The Time Machine. U.S. Copyright has expired; full text available online at Project Gutenberg

Is there some significance to the quote that I'm missing? I'm not catching the relevance to my OP except that both mention that time is (or can be considered as) a dimension.

-FrL-

[Keeve]

People tend to equate "dimension" with "spatial distance". I've always thought this portion of that book does a nice job of illustrating how important and real Duration is, despite its not being spatial. That's all.

[silenus]

I went a totally different direction with this, based on just the OP, without reading all the other responses.

1 second = 30.48 meters

[Frylock]

Originally Posted by silenus

I went a totally different direction with this, based on just the OP, without reading all the other responses.

1 second = 30.48 meters

I hope you mean you based your answer just on the title of the OP.

-FrL-

[silenus]

Well, yeah. I'm on vacation and the wine is sweet...what can I say?

[Mathochist]

Originally Posted by Frylock

Is there an answer to the question in the title of my post? If time is a dimension in the same sense as the spatial dimensions are dimensions, then I might think there should be a sensible way to convert meters into seconds and vice versa.

Well, is there?

-FrL-

I actually work with this sort of thing. Contrary to what Orbifold says, the speed of light in a vacuum is the conversion factor, as other people have said, and there is a physical reason for it.

Let's say you and I are standing on a plane (infinite flat surface). We have two odd compasses, and each one gives a different direction for "north". We want to find out how far a post is from our current position. I use my notion of "north" and "east" to find out how far north and how far east from us the post is, and call those numbers n_M and e_M. You use your notions of "north and "east" to get the coordinates of the post and find numbers n_F and e_F. Since your notion of north and east are different from mine, our coordinates for the post disagree.

Now we each try to calculate the distance using the Pythagorean theorem. I calculate

d_M² = n_M² + e_M²

while you calculate

d_F² = n_F² + e_F²

What do we find? Our calculations agree! It doesn't matter what we each call "north" and "east": our notions of "distance" are the same.

Now when we do this in spacetime we each have a notion of "right", "forward", "up", and "forward in time" (x, y, z, t). If we're at the same point in spacetime and we want to look at some other point, we'll each get different spacetime coordinates for that point. However, there's something that we will agree on: the "spacetime displacement" or "squared-distance" to that point.

d² = x² + y² + z² - c²t²

To bring time displacements into the formula, we have to convert our time measurements (in seconds) to space measurements (in meters). The conversion factor is exactly c. No matter what x, y, z, and t directions you and I pick separately, we will always agree on the result of the above formula.

The weird thing about time is that there's a minus sign in the formula for spacetime displacement rather than a plus sign like for the spatial coordinates. That has deep consequences for the geometry of spacetime, but the formula makes clear that to consider space and time on the same footing we have to use c as our conversion factor.

[lissener]

How many grams in a cup?

See the problem? They're units of measurement; metaphors, when you come down to it; not actual things. In effect they're like adjectives, not nouns. You need a noun to modify, or your adjective has nothing to hang on. You need something to measure, not just the abstract measurements themselves.

[lissener]

It's like you're asking, "How big is red?"

"Big" and "red" only exist in relation to the thing they describe.

[Mathochist]

Originally Posted by lissener

How many grams in a cup?

See the problem? They're units of measurement; metaphors, when you come down to it; not actual things. In effect they're like adjectives, not nouns. You need a noun to modify, or your adjective has nothing to hang on. You need something to measure, not just the abstract measurements themselves.

This is complete nonsense. A measurement is not a metaphor, and it's not an adjective. A meter is a very well defined amount of spatial displacement, and a second is a very well defined amount of temporal displacement.

The reason "how many grams in a cup" doesn't work is that grams measure mass and cups measure volume. Mass is not the same as volume. What Einsteinian relativity tells is is that spatial displacements are the same as temporal displacements, and the conversion factor between a spatial unit and a temporal unit is c, in those units.

How many meters in a second? c, in meters/second. How many miles in an hour? c, in miles/hour. How many light-years in a year? 1.

[robardin]

I have no personal expertise in this area, I just find the question interesting,
but my "life partner" happens to be a mathematician (a tenured research professor in Riemannian Geometry) who was also a Physics minor in college, who had this to say via a terse e-mail reply after I brought the subject up (Orbifold and Mathochist, please don't shoot the messenger) and after scanning the postings so far:

Orbifold is explaining it 100% correctly and is being ignored.

Mathochist is just describing Minkowski spacetime: a model which does assign points a location. Orbifold is describing the reality of spacetime as a manifold (or an orbifold as some people call it). One of the points of general relativity is that there aren't distances between locations, there is no concept of perpendicular between the time dimension and the space dimensions, that two space time points appear to have the same location for one observer and not another.

I suspect Mathochist is a mathematician who is not a geometer and orbifold is a geometer.

[Frylock]

I don't understand how one could say Orbifold is being ignored. But that's a nitpick.

I thought with Mathochist's post I had finally found salvation, but now I believe myself as ignorant as I believed myself before.

Please, everyone, continue your altercation for my amusement.

-FrL-

[Mathochist]

Originally Posted by robardin

I have no personal expertise in this area, I just find the question interesting,
but my "life partner" happens to be a mathematician (a tenured research professor in Riemannian Geometry) who was also a Physics minor in college, who had this to say via a terse e-mail reply after I brought the subject up (Orbifold and Mathochist, please don't shoot the messenger) and after scanning the postings so far:

I am describing Minkowski spacetime because I don't want to complicate the matter with curvature when it's not an essential part of the description. Spacetime is locally Minkowskian, and that's what we're talking about. Orbifold asserts that "there's no physical reason to say that 1 second is 299,792,458 metres." I'm asserting that there is a compelling physical reason. Technically it's involved in measuring the lengths of tangent vectors, which do form a Minkowskian space. I alternately get yelled at for going into too much detail and get yelled at for streamlining my explanations around here. This time it's the latter.

Still, the upshot is that you can't just use any old conversion factor to put time and space on an equal footing in the formula for spacetime displacement. I can't say 1s = 1m and get a sensible result. I can't say 1s = 80m and get a sensible result. Once I choose a unit to measure lengths and one to measure times the conversion factor must be the speed of light in those units.

d² = x² + y² + z² - c²t²

Of course there are other formulas that we'll agree on, as Orbifold mentioned in another post

t² - x²/c² - y²/c² - z²/c²

calculates the square of the "proper time" between two events, as opposed to that of the "spacetime displacement" that I've been talking about. But still we're using the speed of light in a vacuum to convert. This time it's just converting distance to time. You'd have to be a lunatic to say that the conversion factor "3" only converts from feet to yards and can't be used to convert yards to feet. It's the exact same thing, just run backwards.

Oh, and I am well-versed in geometry and theoretical physics, though knot theory and category theory are my stock-in-trade.

[Mathochist]

Incidentally, robardin, I'm fascinated by coincidence. I'll try to probe this without being too public: are you in computer science yourself?