Donne ≠ Marvell, btw.
I should also suggest that your wandering black hole towards the low end of your 10-10,000 solar mass range, since supermassive black holes are very rare and generally found in the centre of galaxies. I would expect most wandering black holes to have a mass somewhere between 4 and 20 solar masses.
:smack:
Thanks. Sorry. But it is cool as a poetic analogue, right?
Thread winer. 
Your velocity with respect to what? According to relativity, velocity is always with respect to some reference frame. If that reference frame was your original one (and you were pretty much still with respect to the rest of the universe), then my post applies. If it’s WRT a bit of stuff you jettisoned before accelerating, it’d be WRT that jetsam, and you’d have to apply the Lorentz transform to calculate your velocity relative to that jetsam.
I say again: velocity is ALWAYS with respect to something. Until you identify that something, velocity is not meaningful. And as I said above, when comparing or adding or subtracting two velocities, if they’re not WRT the same reference frame, you have to apply the Lorentz transformation to transform one of them to the reference frame of the other (or to both to convert them to a 3rd frame).
If they reported that their proper velocity (in some frame) was greater than 1, it wouldn’t provoke any comment (well, it would, because that’s still a honking big speed, but it’s not impossible). If they then converted that proper velocity to a velocity, they’d find that it’s still less than 1. If they reported a velocity greater than 1, then the response would be that they screwed up the calculation.
For manson1972’s benefit –
Chronos is using natural units here, which is why he’s saying “1” and not “c”. Just replace “1” with “c” in the above to bring it back to the conventions used so far in the thread.
I had some time recently to play with this more (and maybe I’ll give a talk about this problem next time I speak at an SF convention - it was fun). If we assume (for simplicity’s sake - I’m just trying to get a rough idea of the magnitudes involved) that the black hole is moving so that when it is closest to the Earth, it is also closest to the Sun (so that that Sun, the Earth, and the black hole are on a straight line at the moment that the black hole is closest to the Earth), and that the black hole is moving fast enough so that the Earth doesn’t move very much relative to the Sun while the black hole is close enough to significantly affect the Sun and the Earth, then the effect of the black hole can be treated as an instantaneous increase in the Earth’s radial velocity, and by using the vis-viva equation (Vis-viva equation - Wikipedia) the change in the Earth’s semimajor axis can be determined.
The increase in the Earth’s radial velocity can be found by integration of the equations I mention in my earlier reply , and if the black hole never gets closer than about 10 au from the Earth, then I can use approximations to get a particularly nice equation using integral tables (God’s gift to the engineer!). Under those circumstances, delta V is piGMh*d/2/l/l/v where G is the gravitational constant, M is the mass of the black hole, d is the distance from the Earth to the Sun, l is the smallest distance from the black hole to the point half way between the Sun and the Earth, and v is the velocity of the black hole. This equation accords with what we would expect - the effect is smaller if the black hole is moving fast and never gets that close to the Earth, and larger the more massive the black hole is and the further apart the Earth and Sun are.
If I’ve done my algebra right, the change in semimajor axis of the Earth is proportional to the square of the velocity impulse, and if we put the mass of the black hole in terms of the Sun’s mass and all the distances in terms of astronomical units, then the velocity impulse (call it delta) equals piMb/(2llv) and the change in semimajor axis is delta^2/(1-delta^2). So if you want the change in semimajor axis to be small enough to not kill everybody (let’s say a change of less than 1%), the black hole has to be really small, really fast, and not particularly close to the inner solar system. Hope that helps.
I noticed an error and an ambiguity. First, the error: It’s the velocity impulse squared (delta^2) that equals piMb/(2llv) , which is great since delta squared that is needed in the next equation. Second, v (the velocity of the black hole) should be in terms of Earth’s orbital velocity (30 km/s). I’ll work an example - a black hole moving at 8000 km/sec as the OP suggested, of mass 10 times the sun. If the closest the black hole gets to the point halfway between the Sun and the Earth is 10 au, then delta squared = pi10/(21010270)=5.8*10^(-4), so that black hole would alter the Earth’s semimajor axis by about 6 ten-thousandths - which is a pretty small effect.
And here’s an article about one Giant Black Hole Kicked Out of Home Galaxy - NASA
“Astronomers have found strong evidence that a massive black hole is being ejected from its host galaxy at a speed of several million miles per hour. New observations from NASA’s Chandra X-ray Observatory suggest that the black hole collided and merged with another black hole and received a powerful recoil kick from gravitational wave radiation.”
If the gravitational radiation is strong enough to propel a black hole to several million miles per hour, what would the radiation do to gas or stars that it hits? Would there be observable effects?
Not unless the matter was already just about to fall into one of the holes, within ten or so radii of the hole. You may have heard about how non-interactive neutrinos are, and how a gazillion of them pass through us every second without us ever noticing, but neutrinos have nothing on gravitational waves.
Thanks, Chronos, for your excellent explanations.
The 4-vector is spacetime. It has three dimensions of space and one of time. The effect is due to the time dimension, but is being applied to one of the space dimensions.
Or, as Chronos said, the proper formula for momentum has a scaling factor, gamma. At really low velocities (i.e. normal human speeds), gamma ~ 1. Ergo, the equation p = m * gamma * v approximates to p = m * v.
To give an analogy, suppose you wanted to measure the time it takes to walk down a long hallway. There are an unspecified number of doors along the hallway that are closed and you must open them to continue. Your time would be the sum of the time walking at your walking speed and the time stopped to open the doors.
t = (L / ws) + (td * N)
where
t = total time
L = hallway length
ws = walking speed
td = time to open 1 door
N = number of doors
For a large number of doors, the time approaches the time it takes to open the doors, and your walking speed becomes negligible. For a fast walking speed, that is also the case.
Now suppose you wished to approximate the time it takes to walk the hallway, and instead of multiplying the number of doors against a “door opening time”, instead you multiplied it against a “hallway length factor”.
L = hallway length
Lo = relativistic hallway length
D = door factor
Then
Lo = L * D * N
And then plug that into your average velocity
t = Lo / ws
That will give you the time it takes to walk the hallway. But it makes it seem like you are making the hallway longer, when what you are doing is inserting more doors.
That’s kinda the same thing. The relativistic effect that prevents you reaching c is not that the mass increases, it is that spacetime is curved, and going faster makes it harder to go through spacetime. Your spacetime gets longer.
My analogy isn’t really what is happening, but shows the kind of mistake that is occurring when using the relativistic mass concept.
Velocity is not measured versus a fixed point, it is measured versus a reference point. You choose the reference point you want to use, but it doesn’t have a privileged position of being “fixed”.
I think Chronos has already explained the concept of relativistic mass and why it is not really helpful.
Measure your acceleration compared to what? You are mixing reference frames. You are measuring against your starting reference point, and then trying to take those numbers and apply them directly in your current reference frame. If you keep your measurements in the same reference frame, you will see consistent results.
You can’t ignore the rest of the world, that is all you have to compare your velocity and acceleration against. That includes your starting point in “empty space”. There may not be a physical object at your “starting point”, but it is still the reference by which you are measuring your velocity and acceleration.
When you state you measure the seconds you accelerated times your acceleration to get your velocity, you are measuring your velocity relative to that initial starting point. You can calculate your time dilation and length contraction in that reference frame.
If you use your current state as your reference frame, you cannot say your velocity is your acceleration times the time. Instead, that is the velocity that your starting point is receding from you. You are still at velocity = 0. The world is accelerating.
Think of this: you have three objects: A, B, C. You can measure the velocity of all three, but you have to state what you are using as your reference. If you use A as your reference, then the velocity of A = 0. You can alternately state those velocities with respect to B, where velocity of B = 0. But those are not the same reference. If the three objects are moving with respect to each other, you will not get both A and B to be zero.
Relativity works the same way. If your reference point is you on the rocket sled, then your velocity will always be zero. Any acceleration you measure will be the acceleration of the rest of the world around you.
Please don’t do that. Strings of division signs are ambiguous.
Much better.
Will do.
Thanks for the great explanation, I understand almost all of what you are saying. But in the quoted sentence, I don’t understand what that means.
If I’m on my rocket sled, fire my rockets and use an accelerometer to measure my acceleration, are you saying, even tho I measure my acceleration, I’M not the thing accelerating?
No, it’s velocity which increases relativistic mass. But anything below .6C is effectively insignificant. 
Even if you accept the notion of “relativistic mass” as meaningful, it’s completely independent of direction. And “effectively insignificant” depends on the effect you’re looking at. You can detect relativistic effects from motion at a speed of literally centimeters per hour, and it isn’t even particularly difficult.
Definition of reference point. The reference point is always zero, and everything is measured with respect to that point. Sure, it’s kinda unintuitive to say I’m at zero and the universe is accelerating away from me, but the math works out cleanly.
I’m sure the math works out. But if I’m holding the accelerometer, and it’s registering acceleration, I don’t get how you can still say “sure, you are measuring acceleration, but you are not really accelerating”
Right now, sitting in your chair, are you accelerating upwards at the 9.8 m/s^2 that an accelerometer would measure? Or are you stationary?
Both.
good question, I don’t know. I don’t have an accelerometer handy to check.