B Time Dilation Problem with Mesons

[NoahsArk]

Thank you for all the helpful responses.

 

[Andrew Mason]

You certainly do not need the Lorentz transformations in this type of problem if you apply the spacetime interval correctly. Even in the case where the meson is moving in both frames you are considering you can relate the times in the frames by using the spacetime interval through $\Delta x = v \Delta t$ and $\Delta x' = v' \Delta t'$. In fact, I would argue that it is much easier than bringing out the Lorentz transformations. Whenever you can use geometrical arguments in relativity (which is most of the time), they tend to be much simpler than looking at coordinates transformations.

And how does one find $\Delta t'$ from v and the proper time without applying at least implicitly using the Lorentz transformation?

AM

 

[Orodruin]

And how does one find $\Delta t'$ from v and the proper time without applying at least implicitly using the Lorentz transformation?

AM

I already told you how to solve the problem of relating the times if you are given the speeds. You will have to be more specific about exactly what problem you want solved.

 

[Andrew Mason]

I already told you how to solve the problem of relating the times if you are given the speeds. You will have to be more specific about exactly what problem you want solved.

Well, my point is that when you use the space-time interval in that way, you are implicitly using the Lorentz transformation:
Using the space-time interval:
\Delta S^2 = c^2t^2 = c^2t^2 - x^2 = c^2t^2 - (vt)^2 = (c^2 - v^2)t^2
t' = t\frac{1}{\sqrt{1-v^2/c^2}} = \gamma t

Which is just a long way of applying the Lorentz transformation to determine t' from the proper time $t' = \gamma (t - 0)$

AM

 

[Orodruin]

Well, my point is that when you use the space-time interval in that way, you are implicitly using the Lorentz transformation:
Using the space-time interval:
\Delta S^2 = c^2t^2 = c^2t^2 - x^2 = c^2t^2 - (vt)^2 = (c^2 - v^2)t^2
t' = t\frac{1}{\sqrt{1-v^2/c^2}} = \gamma t

Which is just a long way of applying the Lorentz transformation to determine t' from the proper time $t' = \gamma (t - 0)$

AM

No, this is not using the Lorentz transformation. It is using the invariance of the spacetime interval, which in my opinion is more of a fundamental concept than Lorentz transformations (which is just a coordinate transformation between some arbitrary Minkowski coordinate frames). Obviously you must get the same result, since the Lorentz transformation is based on the invariance of the spacetime interval and some assumptions regarding the frames it relates. I also disagree that it is the ”long way”. If you want to use the Lorentz transformation you must first derive it from the invariance of the spacetime interval, then you must be careful in considering what is what.

 

[Andrew Mason]

Obviously you must get the same result, since the Lorentz transformation is based on the invariance of the spacetime interval and some assumptions regarding the frames it relates.
I also disagree that it is the ”long way”. If you want to use the Lorentz transformation you must first derive it from the invariance of the spacetime interval, then you must be careful in considering what is what.

Why would you derive the Lorentz transformation from the invariance of ΔS²? The Lorentz transformations came first, after all. Lorentz derived them before Einstein developed Relativity. They follow from the constancy of the speed of light in all frames of reference. The invariance of the space-time interval derives from the Lorentz transformations. Otherwise, how does on show that the space-time interval is invariant?

AM

 

[Orodruin]

Why would you derive the Lorentz transformation from the invariance of ΔS²? The Lorentz transformations came first, after all. Lorentz derived them before Einstein developed Relativity. The invariance of the space-time interval derives from the Lorentz transformations. Otherwise, how does on show that the space-time interval is invariant?

AM

Historically perhaps. But the historical way of developing the theory is not necessarily the most straightforward one or the most elegant to present. After all, we have had over 100 years to develop the theory since Einstein. The invariance of the spacetime interval is geometrically more fundamental al it does not rely on anything else than the geometry of Minkowski space. You can ask the same question of Euclidean geometry: Would you consider the particular form of rotations more or less fundamental than the fact that the distance function between points? I would argue that the latter is much more defining of the geometry of the space - it is after all directly related to the metric tensor whereas you can use any coordinates you wish and get away with it.

Historically relativity was developed in a certain way, but that was gradually hinting at and moving toward a more geometrical view of spacetime. Relativity is ultimately a theory of the geometry of spacetime, regardless of how it developed. You can of course close your eyes to this and go on primarily using Lorentz transformations, but it will obscure the beauty and deeper understanding in my opinion.

 

[Andrew Mason]

Historically perhaps. But the historical way of developing the theory is not necessarily the most straightforward one or the most elegant to present. After all, we have had over 100 years to develop the theory since Einstein. The invariance of the spacetime interval is geometrically more fundamental al it does not rely on anything else than the geometry of Minkowski space. You can ask the same question of Euclidean geometry: Would you consider the particular form of rotations more or less fundamental than the fact that the distance function between points? I would argue that the latter is much more defining of the geometry of the space - it is after all directly related to the metric tensor whereas you can use any coordinates you wish and get away with it.

• Historically relativity was developed in a certain way, but that was gradually hinting at and moving toward a more geometrical view of spacetime. Relativity is ultimately a theory of the geometry of spacetime, regardless of how it developed. You can of course close your eyes to this and go on primarily using Lorentz transformations, but it will obscure the beauty and deeper understanding in my opinion.

That is certainly one perspective. But I am not sure that the invariance of the space-time interval makes anything easier to understand or is is more illuminating or elegant or easier to apply than the Lorentz transformations. The beauty of Relativity, for me, lies in the way it completely changes and complicates our previously held (Newtonian) notions of time and space from a simple, elegant premise (that all the laws of physics are the same in all inertial frames of reference). The Schrodinger equation and Planck's equation are other simple mathematical statements that reveal a much more complicated world than had been previously thought.
AM

 

[Orodruin]

The beauty of Relativity, for me, lies in the way it completely changes and complicates our previously held (Newtonian) notions of time and space from a simple, elegant premise (that all the laws of physics are the same in all inertial frames of reference).

Then you really should try to embrace the geometrical view of what relativity tells us about spacetime and not get hung up on an arbitrary set of coordinate transformations...

I know it does not come easy. I have spent several years teaching both SR and GR at master level and it really needs to grow on you.

 

[robphy]

But I am not sure that the invariance of the space-time interval makes anything easier to understand or is is more illuminating or elegant or easier to apply than the Lorentz transformations.

A geometrical analogy might be helpful here.
Consider a 345-right triangle in the Euclidean plane, with no sides along the x- or y-axes.
Invariance of the distance means we can use 3, 4, and 5 immediately, no matter how the sides are oriented.
We could write the pythagorean theorem, or construct ratios of sides.

However, one can take another approach.
One can write the x- and y-components of each side... and work algebraically from there.
Alternatively,
one could rotate the triangle so that one side is aligned with one of the x- or y-axes, then measure the length of that side.
If necessary, rotate the triangle again into some "standard position" to get the length of one or more of the other sides.