- #1

- 580

- 21

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- I
- Thread starter Frank Castle
- Start date

- #1

- 580

- 21

- #2

- 20,125

- 10,865

$$\frac{\mathrm{d} x^{\mu}}{\mathrm{d} \lambda} \frac{\mathrm{d} x_{\mu}}{\mathrm{d} \lambda}>0.$$

Here ##\lambda## is an arbitrary parameter for the world line. You can take the coordinate time ##t=x^0## for it.

Thus proper time

$$\tau=\int_{0}^t \mathrm{d} \tilde{t} \sqrt{\mathrm{d} x^{\mu}/\mathrm{d} \tilde{t} \mathrm{d} x_{\mu}/\mathrm{d} \tilde{t}}$$

is a monotonously increding function of ##t##, and thus you can prametrize the world line as well with ##\tau## as you can parametrize it with ##t##.

For a curve that is light-like you have ##\tau=\text{const}##, i.e., there's no proper time for light-like world lines. In other words, massless particles cannot be parametrized with any kind of "proper time". You have to take arbitrary other parameters, including the coordinate time of an arbitrary inertial frame.

For space-like curves all of this doesn't make any sense at all. Tachyons are very tricky classically and impossible to define as interacting "particles" in the quantum case. You can forget about them, because up to now there's fotunately no need for them.

- #3

- 580

- 21

You can take the coordinate time t=x0t=x^0 for it.

How can one use coordinate time as a parameter when it isn't Lorentz invariant?

is a monotonously increding function of tt, and thus you can prametrize the world line as well with τ\tau as you can parametrize it with tt.

So is an object required to be monotonically increasing in order for it to be a parameter?

- #4

PAllen

Science Advisor

- 8,851

- 2,051

A parameter, in general, is arbitrary, not observable, and need not be invariant. See next answer.How can one use coordinate time as a parameter when it isn't Lorentz invariant?

A parameter is just an interval of real numbers that is mapped continuously to points on a curve. The mapping must be a bijection. Given a parametrization of a curve, any monotonically increasing function of that parameter is also a valid parameter (that preserves the 'direction' of mapping).So is an object required to be monotonically increasing in order for it to be a parameter?

- #5

- 580

- 21

A parameter, in general, is arbitrary, not observable, and need not be invariant. See next answer.

A parameter is just an interval of real numbers that is mapped continuously to points on a curve. The mapping must be a bijection. Given a parametrization of a curve, any monotonically increasing function of that parameter is also a valid parameter (that preserves the 'direction' of mapping).

Ah ok, that makes sense. I think what confuses me in this case is the fact that coordinate time is observable in special relativity and so it blurs the lines a little.

- #6

Nugatory

Mentor

- 13,856

- 7,250

Ah ok, that makes sense. I think what confuses me in this case is the fact that coordinate time is observable in special relativity and so it blurs the lines a little.

Even in SR, it's proper time that is the observable. We generally choose the coordinate system so that the time coordinate is equal to the proper time along the worldline of an observer at rest in that coordinate system, but that's still an arbitrary coordinate choice.

- #7

- 8,942

- 2,931

How can one use coordinate time as a parameter when it isn't Lorentz invariant?

Maybe it would help to think of it this way: In parametrizing a curve, you can use an arbitrary real-valued parameter that smoothly increases as you move along the curve. That parameter may as well be

- #8

- 580

- 21

Maybe it would help to think of it this way: In parametrizing a curve, you can use an arbitrary real-valued parameter that smoothly increases as you move along the curve. That parameter may as well beSOMEBODY'scoordinate time. Because coordinate time is not Lorentz invariant, it won't be coordinate time for everyone, but that doesn't matter. Everyone can agree that it is a quantity that increases along the curve, even if everyone doesn't agree that it's coordinate time.

Thanks, this makes sense to me. Is it correct to say that as the curve is timelike the quantity will increase in everyone's reference frame, even if it doesn't correspond to their coordinate time?

- #9

PAllen

Science Advisor

- 8,851

- 2,051

You can parametrize arbitrary curves, even ones that vary between timelike, null, and spacelike along their path. The parameter will be functionally related to proper time for a timelike curve; proper length for a spacelike curve; and not related to anything meaningful for an arbitrary curve. Obviously, an arbitrary curve has no physical meaning, but is still a mathematical curver.Thanks, this makes sense to me. Is it correct to say that as the curve is timelike the quantity will increase in everyone's reference frame, even if it doesn't correspond to their coordinate time?

- #10

PAllen

Science Advisor

- 8,851

- 2,051

Thanks, this makes sense to me. Is it correct to say that as the curve is timelike the quantity will increase in everyone's reference frame, even if it doesn't correspond to their coordinate time?

- #11

- 8,942

- 2,931

Thanks, this makes sense to me. Is it correct to say that as the curve is timelike the quantity will increase in everyone's reference frame, even if it doesn't correspond to their coordinate time?

Yes. So you can't parametrize the curve by [itex]x[/itex], for instance, because [itex]x[/itex] won't necessarily increase all along the curve.

- #12

- 580

- 21

Yes. So you can't parametrize the curve by xx, for instance, because xx won't necessarily increase all along the curve.

Would it be correct to say that the arc-length of a timelike curve is equal to the elapsed proper time along that curve and so we are simply parametrising the curve by its arc-length?

- #13

- 8,942

- 2,931

Would it be correct to say that the arc-length of a timelike curve is equal to the elapsed proper time along that curve and so we are simply parametrising the curve by its arc-length?

Yes, if the curve is timelike.

- #14

- 580

- 21

Yes, if the curve is timelike.

Ok, cool. I think from this (for some reason) more than anything, it makes sense to me why we can use proper time as a parametrisation for a timelike curve.

Thanks for your help!

- #15

- 20,125

- 10,865

If there's any general message of relativity it's that you have to be careful about how you define an observable, and you observe the quantity you measure. E.g., the famous example about the atmospheric muons reaching the earth although from a naive point of view there shouldn't be many left in the naive Newtonian view of their "lifetime" due to relativistic time dilation tells you that here you measure the lifetime of a fast-moving particle in the (coordinate) time of the observer at rest with respect to earth. Of course, what's listed in the particle data booklet as the lifetime of the muon is by definition (!) that of a muon at rest, i.e., in it's proper frame. So it's not always this proper time that's relevant to describe an observed phenomenon, i.e., here the prolongued lifetime due to relativistic time dilation explaining why so much more muons reach the earth compared to the naive (and wrong!) expectation based on the (wrong!) use of the Newtonian space-time model.Even in SR, it's proper time that is the observable. We generally choose the coordinate system so that the time coordinate is equal to the proper time along the worldline of an observer at rest in that coordinate system, but that's still an arbitrary coordinate choice.

- #16

PAllen

Science Advisor

- 8,851

- 2,051

But the proper time elapsed for the muons is much less than rest half life, and that is fully equivalent to the literal observation of survival rate. Survival rate and half life => proper time elapsed. There are then two coordinated dependent explanations: time dilation in the earth frame, length contraction of the atmosphere in the muon frame. In either case, the literal observable is equivalent to the invariant proper time.If there's any general message of relativity it's that you have to be careful about how you define an observable, and you observe the quantity you measure. E.g., the famous example about the atmospheric muons reaching the earth although from a naive point of view there shouldn't be many left in the naive Newtonian view of their "lifetime" due to relativistic time dilation tells you that here you measure the lifetime of a fast-moving particle in the (coordinate) time of the observer at rest with respect to earth. Of course, what's listed in the particle data booklet as the lifetime of the muon is by definition (!) that of a muon at rest, i.e., in it's proper frame. So it's not always this proper time that's relevant to describe an observed phenomenon, i.e., here the prolongued lifetime due to relativistic time dilation explaining why so much more muons reach the earth compared to the naive (and wrong!) expectation based on the (wrong!) use of the Newtonian space-time model.

Share: