B Another question about relativity: the force

[physics user1]

So, i want to derive the relativistic force using the definition F= dp/dt and p= γ mv

I consider a frame of reference solidal with the particle. Algebrically i can do it and it works, but why can i do it? Why do i have to differentiate also γ? That would mean that the speed of the frame of reference changes in time, (and it does, the particle accelerates) but then is no more an inertial frame so how can special relativity hold? Please help me i cannot get out of this problem

 

[Ibix]

What does "solidal" mean?

You are accelerating the particle, so you are changing its speed as measured in any inertial frame. Pick the inertial frame in which the particle is initially moving at v and work in that one. You have the correct formula for the three-momentum, so just differentiate the expression. Why would you leave a term out?

Note you need to differentiate with respect to the particle's proper time, not coordinate time.

 

[physics user1]

What does "solidal" mean?

You are accelerating the particle, so you are changing its speed as measured in any inertial frame. Pick the inertial frame in which the particle is initially moving at v and work in that one. You have the correct formula for the three-momentum, so just differentiate the expression. Why would you leave a term out?

Note you need to differentiate with respect to the particle's proper time, not coordinate time.

Why the particle proper time?

 

[Ibix]

Why the particle proper time?

Because then you're differentiating something that behaves like (part of) a vector with respect to something invariant. You come out with something that behaves like (part of) a vector under Lorentz transforms.

 

[DrGreg]

What does "solidal" mean?

@Cozma Alex, you didn't answer this. I've never seen this word before, and I can't find any definition via Google.

 

[physics user1]

@Cozma Alex, you didn't answer this. I've never seen this word before, and I can't find any definition via Google.

That moves with the particle, a frame where the speed of the particle is 0

 

[DrGreg]

That moves with the particle, a frame where the speed of the particle is 0

I would call that the "rest frame" of the particle.

 

[physics user1]

I would call that the "rest frame" of the particle.

Then i mean the rest frame, if i consider the rest frame, this is accelerating, so its no more inertial, so why can i apply the laws of special relativity here?

 

[pervect]

So, i want to derive the relativistic force using the definition F= dp/dt and p= γ mv

I consider a frame of reference solidal with the particle. Algebrically i can do it and it works, but why can i do it? Why do i have to differentiate also γ? That would mean that the speed of the frame of reference changes in time, (and it does, the particle accelerates) but then is no more an inertial frame so how can special relativity hold? Please help me i cannot get out of this problem

If you have a frame of reference that is instantaneously co-moving with some particle, then the instantaneous velocity of that particle in that instantaneously co-moving inertial frame is zero at some time t=t0. At the times $t0-\delta t$ and $t0 + \delta t$ the velocity of the particle is not necessarily zero, and if there is a force acting on it it will not be zero. But it will be very small.

We can linearly approximate the velocity of the particle near t = t0 as v = 0 + a(t-t0), where a is some constant. So we can approximate the momentum p near t=t0 as p = 0 + ma(t - t0). And find dp/dt = ma.

Because we have chosen an instantaneoudly co-moving frame, we can approximate $\gamma$ as being equal to 1, as v is zero at t=t0, and very small for times close to t=t0.

But this is only true in the instantaneously co-moving frame. If we want to find p in some other frame, we need to use the chain rule, which is what I think you were talking about earlier.With $p = \gamma m v$, $dp/dt = \gamma m (dv/dt) + m v (d/dt) \gamma$ When we evaluate $(d/dt) \gamma$ in an instantaneously co-moving frame, we find that this term is zero, but this is not true in general.

$\frac{d \gamma}{dt} = \frac{d \gamma}{dv} \frac{dv}{dt}$

Assuming c=1 for simplicity (feel free to redo the calculation if you're not comfortable with this assumption, which is equivalent to using gemoetric units) we can write $\gamma = 1/\sqrt{1-v^2}$ and

$$\frac{d\gamma}{dv} = \frac{v}{(1-v^2)^{1.5}}$$

so we see that when v=0, $d\gamma/dv$ = 0 and thus $d\gamma/dt = 0$.

If v is not equal to zero, then this term is NOT zero and must be included.

 

[physics user1]

If you have a frame of reference that is instantaneously co-moving with some particle, then the instantaneous velocity of that particle in that instantaneously co-moving inertial frame is zero at some time t=t0. At the times $t0-\delta t$ and $t0 + \delta t$ the velocity of the particle is not necessarily zero, and if there is a force acting on it it will not be zero. But it will be very small.

We can linearly approximate the velocity of the particle near t = t0 as v = 0 + a(t-t0), where a is some constant. So we can approximate the momentum p near t=t0 as p = 0 + ma(t - t0). And find dp/dt = ma.

Because we have chosen an instantaneoudly co-moving frame, we can approximate $\gamma$ as being equal to 1, as v is zero at t=t0, and very small for times close to t=t0.

But this is only true in the instantaneously co-moving frame. If we want to find p in some other frame, we need to use the chain rule, which is what I think you were talking about earlier.With $p = \gamma m v$, $dp/dt = \gamma m (dv/dt) + m v (d/dt) \gamma$ When we evaluate $(d/dt) \gamma$ in an instantaneously co-moving frame, we find that this term is zero, but this is not true in general.

$\frac{d \gamma}{dt} = \frac{d \gamma}{dv} \frac{dv}{dt}$

Assuming c=1 for simplicity (feel free to redo the calculation if you're not comfortable with this assumption, which is equivalent to using gemoetric units) we can write $\gamma = 1/\sqrt{1-v^2}$ and

$$\frac{d\gamma}{dv} = \frac{v}{(1-v^2)^{1.5}}$$

so we see that when v=0, $d\gamma/dv$ = 0 and thus $d\gamma/dt = 0$.

If v is not equal to zero, then this term is NOT zero and must be included.

Thanks

 

[Dale]

Why the particle proper time?

Because then the resulting quantity is invariant.

 

[physics user1]

Because then the resulting quantity is invariant.

But... we want the particle momentum according to our frame of reference, in our frame of reference to get the velocity of the particle we make dx/dt using our dx and dt, why in this case the velocity is different? Why do be use the same dx but we use instead of dt, dt'?

 

[SiennaTheGr8]

It depends on what you're trying to do.

There's nothing "wrong" with the ordinary time-derivative of three-momentum. It just isn't the (spatial) component of a four-vector. To make it so, scale it by the Lorentz factor (which is equivalent to taking the proper-time–derivative of momentum in the first place).

 

[Ibix]

But... we want the particle momentum according to our frame of reference, in our frame of reference to get the velocity of the particle we make dx/dt using our dx and dt, why in this case the velocity is different? Why do be use the same dx but we use instead of dt, dt'?

Basically, $v=(1,\partial x/\partial t, \partial y/\partial t, \partial z/\partial t)$ is not a four vector but
$V=(\partial t/\partial\tau,\partial x/\partial \tau, \partial y/\partial \tau, \partial z/\partial \tau)$ is. That is, you will find that, $v'\neq\Lambda v$, but $V'=\Lambda V$, where $\Lambda$ represents the Lorentz transform and the primed versions of $v$ and $V$ are just the same as the unprimed versions but using primed coordinates.

The components of $V$ and $V'$ are different, but they represent the same vector and give you the energy and three-momentum measured in your frame (if you multiply by mass). But $v$ and $v'$ are different things, not just different ways of describing the same thing. They are related to the four velocity, but related in different ways in different frames. Which is not helpful.

 

[Dale]

we want the particle momentum according to our frame of reference,

Why would we necessarily want that? Our frame is just as arbitrary as any other frame. It is better to have quantities that are invariant, so they work the same in any frame. Then we can use any frame that is convenient: our frame, their frame, the Earth's frame, the Sun's frame, some random non-inertial frame ...

 

[pervect]

I haven't seen the whole original problem - it'd be easier if it wasn't in a PDF file that I had to download & attempt to virus check. But I can say that I've never seen a textbook discussion of "force" in GR. The simplest approach to teaching GR is to not use force, as it isn't really relevant to the theory.

So if we want to address the concept of "force" at all, this leaves us with two options - one option is to simply say that there is no such thing as "force" in GR, and to further explain the reasoning, which is that the transformation properties of the mathematical descriptoins that are candidates for "forces" aren't the same as real forces, having more similarity to "fictitious forces". The technical name for these not-forces is "Christoffel Symbols", and they're usually talked about at a graduate level.

The other option is to agree on some specific measurement in some particular frame or coordinate system - the result won't have the transformation properties that make it generally viewable as a "force", but we can come up with specific and perhaps useful descriptions of what is going on in some specific circumstances, even if we need to be very cautious about generalizing it to other frames of reference.

A simpler problem that I have thought about & posted about some illustrates some of the pitfalls. Suppose you have an elevator with a "flat" floor, accelerating out in space. And we want to compare the weight of two objects on the floor, one of which is stationary, and one of which is moving.

Well, we could take a couple of approaches, one approach measures the pressure and area of the moving object on the floor, and uses that pressure to "weight" the object, just like we use truck scales to weigh moving trucks. Mathematically we integrate the pressure over the contact area. It's probably not technically feasible to actually build a truck scale that could work at those speeds, but we can still do the thought experiment.

And if we agree on that approach (which might or might not happen), then we can come up with some value, which I would believe would be that the scale reads a factor of gamma higher for the moving mass.

Then we repeat the question - what if we switch to a frame that's comoving with the moving mass? Because we aren't dealing with true forces, the problem is not straightforwards. We need to redo the whole analysis, rather than simply transform the results from one frame to another, as we would and could do if the force were a "real" force.

The approach I favored, which was based on a textbook example of "the proper reference frame of an accelerated observer", wound up with some interesting an unexpected results. The floor turned out not to be flat in this frame, but curved.

One physical effect implied from this image is that a gyroscope attached to the sliding block will rotate. This tuns out to actually happen, and is a not-particularly intuitive result from special relativity called "Thomas Precession".

There is a paper about the Thomas Precession aspect of the problem that is of some interest, as it also shows the floor curving in a momentarily co-moving inertial frame (MCIRF). It also uses the concept of a MCIRF, something that not everyone wanted to do. The published paper was:. http://arxiv.org/abs/0708.2490v1[/PLAIN]

But while interesting, this paper doesn't address the question of "weighing" the moving block, though it does a good job of explaining why a gyroscope attached to the moving block appears to rotate (though it's a somewhat advanced paper).

Note that all of this takes place in flat space-time, and avoids the complexities in the original problem which isn't in flat space-time. Which doesn't make the problem any simpler, it just means that the original problem was even harder than this one.