Why, exactly, in terms I could understand, do things shrink and get more massive as it approaches c? This seems counter-intuitive to me, since if something’s going really fast, it seems that it would blur in the FoR…
Things do not get more massive as they approach the speed of light. To see this imagine yourself traveling along with the object–would you see it increase in mass?
He probably meant that it would get more dense…just a guess.
This would be the special case of relativity to which you refer.
The foreshortening in the direction of travel is called the Lorenz Transformation.
And as far as I know, the closer to the speed of light that you travel, the greater your mass increases.
Here’s a calculator and a brief explanation.
http://www.1728.com/reltivty.htm
Don’t forget that as you approach the speed of light, time slows down for you. (Called time dilation)
All three of these cases are in reference RELATIVE to a stationary observer.
(hence the term relativity).
I think more SDMB members will jump in here.
The idea that the mass of particles increases as their velocity approaches c used to be how physicists thought about relativity, but now the notion is viewed as somewhat quaint (at best.) It’s true that as a particle goes faster, the momentum increases faster than one would expect; instead of p = m[sub]0[/sub]v (which is what we expect from regular old Newtonian mechanics), we have
p = m[sub]0[/sub] [1/sqrt(1 - v[sup]2[/sup]/c[sup]2[/sup])] v.
The factor in square brackets above is approximately 1 for values of v much less than c, but it starts to increase as v gets to be a significant fraction of the speed of light, and in fact becomes infinite as v approaches c.
There are two ways to interpret this equation. The first is to say that the mass of the particle depends on its velocity, i.e.
m(v) = m[sub]0[/sub]/sqrt(1 - v[sup]2[/sup]/c[sup]2[/sup])
Under this intepretation, the mass of the particle does indeed get larger as its velocity increases.
The other way to think about it, and the way most physicists have thought about it since the 60s or so, is that the “velocity” that we know & love from our daily lives is really only an approximation to some more general thing called the “four-velocity”. Instead of p = m[sub]0[/sub]v, we have p = m[sub]0[/sub]u, where
u = v/sqrt(1 - v[sup]2[/sup]/c[sup]2[/sup])
This may seem to be splitting hairs, but the whole notion of using “four-vectors” instead of vectors is a much more elegant (to physicists) way of thinking about special relativity. If you’re interested, I can try to explain it further.
(If, on the other hand, I’ve completely lost you in the above explanation, let me know & I’ll try to explain it another way.)
Here’s the thing: nothing really changes. Let’s say you and a friend take a picture of a cube, you looking directly at one face and he turned and looking at the corner. The pictures look different, but nothing’s really changed about the cube between the two of you.
Now take a cube one meter on a side and made of such a material that it masses exactly one kilogram. Fly it past you at a constant velocity and measure its side lengths and mass at the instant it passes. You’ll find the side in the direction it’s flying looks shorter than a meter while the other sides (perpendicular to this direction) are still a meter each. The mass will seem to be more than a kilogram. Both of these are just how the cube looks to you, rather than anything that’s changed about the cube itself. Your friend travelling with the cube and making the same measurements at the exact same times as you do will still get a meter on all sides and a mass of exactly one kilogram.
:smack: I didn’t say this, but I did understand the whole “as perceived by others thing” already. I just needed some sort of explanation as to why that happens… or is that too complicated, and I should just assume it’s true?
My best explanation is that it’s pretty much the exact same thing as what happens when you rotate your point of view in space. There’s a more technical explanation, but this way lies math: be warned.
In three dimensional space, we can measure the distance from a point (we’ll call it “the origin”) to any other by picking an orthonormal coordinate system. That is, pick three mutually perpendicular vectors e[sub]1[/sub], e[sub]2[/sub], e[sub]3[/sub] of unit length pointing out from the origin. Now, take the vector from the origin to the other point and write it as
v = x[sub]1[/sub]e[sub]1[/sub] + x[sub]2[/sub]e[sub]2[/sub] + x[sub]3[/sub]e[sub]3[/sub]
By the Pythagorean theorem, the distance is
||v||[sup]2[/sup] = x[sub]1[/sub][sup]2[/sup] + x[sub]2[/sub][sup]2[/sup] + x[sub]3[/sub][sup]2[/sup]
Now, rotating your point of view is the same as picking a different orthonormal system f[sub]1[/sub], f[sub]2[/sub], f[sub]3[/sub]. We can write
f[sub]1[/sub] = a[sub]11[/sub]e[sub]1[/sub] + a[sub]12[/sub]e[sub]2[/sub] + a[sub]13[/sub]e[sub]3[/sub]
f[sub]2[/sub] = a[sub]21[/sub]e[sub]1[/sub] + a[sub]22[/sub]e[sub]2[/sub] + a[sub]23[/sub]e[sub]3[/sub]
f[sub]3[/sub] = a[sub]31[/sub]e[sub]1[/sub] + a[sub]32[/sub]e[sub]2[/sub] + a[sub]33[/sub]e[sub]3[/sub]
Linear algebra tells us that the matrix A of these a[sub]ij[/sub] is orthogonal. That is, its transpose is its inverse.
Now we can rewrite the displacement we’re measuring as
v = y[sub]1[/sub]f[sub]1[/sub] + y[sub]2[/sub]f[sub]2[/sub] + y[sub]3[/sub]f[sub]3[/sub]
Now, we can substitute our expressions for f[sub]i[/sub] into this formula to get
v = y[sub]1[/sub](a[sub]11[/sub]e[sub]1[/sub] + a[sub]12[/sub]e[sub]2[/sub] + a[sub]13[/sub]e[sub]3[/sub]) + y[sub]2[/sub](a[sub]21[/sub]e[sub]1[/sub] + a[sub]22[/sub]e[sub]2[/sub] + a[sub]23[/sub]e[sub]3[/sub]) + y[sub]3[/sub](a[sub]31[/sub]e[sub]1[/sub] + a[sub]32[/sub]e[sub]2[/sub] + a[sub]33[/sub]e[sub]3[/sub]) = (y[sub]1[/sub]a[sub]11[/sub] + y[sub]2[/sub]a[sub]21[/sub] +y[sub]3[/sub]a[sub]31[/sub])e[sub]1[/sub] + (y[sub]1[/sub]a[sub]12[/sub] + y[sub]2[/sub]a[sub]22[/sub] +y[sub]3[/sub]a[sub]32[/sub])e[sub]2[/sub] + (y[sub]1[/sub]a[sub]13[/sub] + y[sub]2[/sub]a[sub]23[/sub] +y[sub]3[/sub]a[sub]33[/sub])e[sub]3[/sub]
Now these coefficients of e[sub]i[/sub] must be the same x[sub]i[/sub] as before, so we plug them into the expression for the length
||v||[sup]2[/sup] = (y[sub]1[/sub]a[sub]11[/sub] + y[sub]2[/sub]a[sub]21[/sub] +y[sub]3[/sub]a[sub]31[/sub])[sup]2[/sup] + (y[sub]1[/sub]a[sub]12[/sub] + y[sub]2[/sub]a[sub]22[/sub] +y[sub]3[/sub]a[sub]32[/sub])[sup]2[/sup] + (y[sub]1[/sub]a[sub]13[/sub] + y[sub]2[/sub]a[sub]23[/sub] +y[sub]3[/sub]a[sub]33[/sub])[sup]2[/sup]
Now, expanding this out in terms of the y[sub]i[/sub] and using the orthogonality properties of A, we find this is exactly
y[sub]1[/sub][sup]2[/sup] + y[sub]2[/sub][sup]2[/sup] + y[sub]3[/sub][sup]2[/sup]
That is, the Pythagorean distance formula holds no matter which orthonormal frame we pick.
Let’s imagine we didn’t know anything about the Pythagorean formula. We talk about displacements in terms of coordinate systems alone. Suddenly, someone notices that there’s this formula that stays the same no matter which coordinate frame we pick. The “distance” exists without picking any coordinates. This seems pretty obvious to you and me, but it’s almost exactly what happened with spacetime.
There are two obstacles, though. One is that we use a different system of units for different directions. Measurements in space are done in meters, while measurements in time are done in seconds. Secondly is that the formula looks slightly different. Someone noticed, though, that if two different people set up measuring systems and they measure the coordinate displacements between two different points in spacetime. One of them gets (x,y,z,t) and the other gets (x’,y’,z’,t’). Now, someone noticed that
x[sup]2[/sup] + y[sup]2[/sup] + z[sup]2[/sup] - c[sup]2[/sup]t[sup]2[/sup] = x’[sup]2[/sup] + y’[sup]2[/sup] + z’[sup]2[/sup] - c[sup]2[/sup]t’[sup]2[/sup]
That is, c (the speed of light in meters per second) is a conversion factor from measurements in seconds to measurements in meters and then applying something like the Pythagorean formula (with a negative sign on time measurements) gives a quantity that doesn’t depend on the orthonormal coordinate system used in spacetime. This is called the “spacetime interval”.
Now, the collection of transformations in three dimensional space which left distances invariant is the “orthogonal group” of R[sup]3[/sup], or O(3). Similarly, in R[sup]4[/sup] we can define the group of transformations leaving the spacetime interval invariant: O(3,1), or the “Lorentz group”. Now, just as there are elements of O(3) which describe rotating one’s reference frame in the x-y plane, there are elements of O(3,1) which describe rotating one’s reference frame in the x-t plane. Two observers whose reference frames differ by such a rotation will look to each other to be moving with a uniform velocity in the x direction.
Now, everything else in special relativity comes out of the properties of this four-dimensional viewpoint. An interval in spacetime exists independantly of the observer’s coordinates, but different observers break that interval into a “space” displacement and a “time” displacement differently. Similarly, the mass of an object turns out to be the “time component” of its four-dimensional momentum. One observer sees the object at rest with mass m, giving four-momentum (0,0,0,m) while the other sees it moving with four-momentum (p[sub]x[/sub],p[sub]y[/sub],p[sub]z[/sub],m’) in such a way that
p[sub]x[/sub][sup]2[/sup] + p[sub]y[/sub][sup]2[/sup] + p[sub]z[/sub][sup]2[/sup] - m’[sup]2[/sup] = -m[sup]2[/sup]
from which it’s easily seen that a nonzero three-momentum implies an increase in mass to balance out.
Wow, and I thought my explanation was complicated. 
Ok I’m confused. If I’m not mistaken the time component of the four-momentum is energy not mass. Mass is the magnitude of the four-vector and by definition it’s invariant. In other words in either frame the mass remains the same, but the energy and momentum differ.
If this is incorrect please explain.
What, you don’t remember the most famous equation of all relativity theory? E=mc[sup]2[/sup]! Energy and mass are the same thing. Just like one meter is c seconds, one Joule (kg m[sup]2[/sup]/s[sup]2[/sup]) is c[sup]2[/sup] kilograms. In both cases c is a conversion factor.
Now, the squared-length of the 4-momentum is Lorentz invariant and is -m[sup]2[/sup] in the rest frame of a massive particle of mass-energy m, however the observed mass-energy is the “time component” of the four-momentum which is not invariant. Remember, as I stated in my first post (which chaoticdonkey then said he already knew so I didn’t mention it again) all the “changes” in special relativity are only apparent changes, so when I say “an increase in mass” I mean it as shorthand for “an increase in the mass as measured by an observer in uniform rectilinear motion with respect to the particle”.
This is the more modern way of thinking about things. In general, the four-momentum p is given in terms of the four-velocity u by p = mu. But the four-velocity is given by
u = [symbol]g[/symbol] (v[sub]x[/sub], v[sub]y[/sub], v[sub]z[/sub], c)
where [symbol]g[/symbol] = 1/sqrt(1 - v[sup]2[/sup]/c[sup]2[/sup]). It’s not too hard to show that the magnitude-squared of this vector (as Mathochist has defined it) is -c[sup]2[/sup]; so the magnitude-squared of p is always -m[sup]2[/sup]c[sup]2[/sup].
It’s possible, though, to view this another way: we can write
p = m[symbol]g[/symbol] (v[sub]x[/sub], v[sub]y[/sub], v[sub]z[/sub], c)
and (as I said before) view m[symbol]g[/symbol] as some sort of velocity-dependent mass. Under this interpretation, we always have E = m(v)c[sup]2[/sup] and p = m(v) v; it’s just that an observer moving relative to the particle will measure a different “mass” from an observer at rest relative to the particle.
Like I said, though, this way of looking at things is somewhat antiquated.
(Oh, and Mathochist: I’m glad you’re using the proper ++± signature for the metric instead of —+, but dammit, the time-component of a four-vector is supposed to be the first one, not the last…)
That equation has limited application the full equation is
m[sup]2[/sup] = E[sup]2[/sup] - p[sup]2[/sup]
(c=1)
Energy and mass are not the same thing. They’re different propetries of a system. A single photon has energy but no mass. A system of photons with a center of momentum frame has both energy and mass. Obviously they can’t be the same thing.
<snip>
I agree
I agree that the time component of the four momentum is not invariant but the time component is energy not mass or mass energy. As shown above they are not the same thing.
Read MikeS’s first post. What you are referring to is relativistic mass which is an outdated concept and which can be transformed away. I’m not exactly sure what you mean by apparent changes but these changes have very real physical ramifications such as electromagnetism.
Disclaimer: this amounts to a hijack to respond to MikeS.
Actually, it doesn’t really matter which you pick within relativity theory. O(3,1) is isomorphic to O(1,3). I could see it becoming a problem when adding quantum field theory since the Clifford algebras of O(3,1) and O(1,3) are distinct, but I believe Spin(3,1) and Spin(1,3) are again isomorphic.
As for the ordering, yes I know that the odd sign out is usually in the zero slot and the fourth slot is more often used for Wick-rotated time coordinates. Aside: yes, Wick rotation seemed rather archaic to me the first few times I saw it, but when you start calculating quantum tunneling effects on instantons it’s nice to have a Riemannian framework rather than merely semi-Riemannian. Anyhow, I suppose I also “should” have written the components of vectors with superscript indices and employed the Einstein summation convention, that being the more modern (and useful) approach.
So, why did I make these choices for my post? Pedagogy. Yes, the poster to which I was directly applying probably could follow easily throwing time in the zero slot, but I was trying to write for anyone who might glance across this thread. That’s why I wrote out the huge reexplanation of distance in R[sup]3[/sup]. The mathematically naive just plain accept more easily the new variable being tacked onto the end rather than shoved in the beginning. Similarly, my sign convention (though I agree with you) is the closest in analogy to the Riemannian case. I could say to add in time with a negative in the metric, or to add it and then flip all the spatial entries in the metric. The former goes down a lot more easily. Add to this the fact that the more mathematically and physically savvy already understand and will see exactly what I’m doing and compensate for it and you get an increase in communication to the low end while sacrificing nothing on the high end.
I will give you this: if someone reads my post and then goes off to a “real” physics course, they will have to learn a different convention than the one I used. Even this, though, can be useful. If they see the same thing written different ways they might start to see the concept and not the notation.
Then again, if they’re in a physics class that’s pretty unlikely (damned index jugglers) 
Just so I know…the four-vector is used because of four dimensional spacetime?
I think quite a bit of what you see as disagreement comes from your more orthodox physical view and mine as a mathematician who plays with this “real world” toy from time to time and amounts to different terminology. Except this point.
By “apparent changes”, I mean that the changes are only apparent. The mass of the particle doesn’t “really” increase, but it seems to increase as the coordinate frames are rotates. I must protest your saying that these changes have real physical effects, though. The principle of general covariance (Lorentz invariance for SR) states – nay demands – that the only thing that has any effect is the geometric object defined independantly of all coordinate frames. That is, the momentum covector (the gloves are off, btw, and I’m distinguishing the tangent and cotangent bundles) exists without any observers around to put coordinate frames on it. The zero component is an artifact of measurement in a coordinate system and can have no physical meaning separate from the other components and the coordinate system being used.
If you mean that when rotating a given momentum in a certain electromagnetic field, the “increased” mass has an effect, you’re absolutely correct. However, I’m emphasizing coordinate changes on a fixed system. If you rotate the momentum in this view you must change the whole coordinate system, thus changing the expression of the field in local coordinates. You simply cannot do calculations with the particle in one coordinate frame and the field in another.
Maybe it just got out of the pool?
First, to respond to Enola Straight: Yup.
(now continuing the sidebar discussion with Mathochist)
No worries, no worries. It’s purely a matter of convention, and pedagogy is always an important consideration. It’s just a reflex action on my part: in the physics department where I work, particle physicists outnumber relativists by about 3 to 1; so us relativists tend to be very protective of our sign conventions.
Easy answer: yes
I can add anything to the math/physics here, but I can add a geeky limerick.
There was a young racer named Fiske
Who took a considerable risk
When his dragster caught traction
The Fitzgerald Contraction*
Reduced his wazoo to a disk!
*another name for the Lorentz Tranformation
