Time Dilation Problem with Mesons

[NoahsArk]

At the suggestion of some of the members here, I am reading the book Space-Time Physics by Wheeler. The last problem in the first chapter states:

"In a given sample of mesons, half will decay in 18 nanoseconds (18 x 10^-9)), measured in a reference frame in which the mesons are at rest. Half of them will decay in the next 18 nanoseconds, and so on.
a) In a particle accelerator mesons are produced when a proton beam strikes an aluminum target inside the accelerator. Mesons leave this target with nearly the speed of light. If there were no time stretching and if no mesons were removed from the resulting beam by collisions, what would be the greatest distance from the target at which half of the mesons would remain undecayed.
b) The mesons of interest in a particular experiment have a speed of .9978 that of light. By what factor is the predicted distance from the target for half-decay increased by the time dilation over the previous prediction- that is, by what factor does the dilation effect allow one to increase the separation between the detecting equipment and target."

For the answer to part a) (which is also given in the book but without explanation), I get 5.4 meters. I did this by calculating 18 nanoseconds by .299972458 meters which is the distance light travels in one nanosecond.

As the answer to part b) (which is not given), I get a factor of 11.74 because the particles can travel 11.74 times farther due to the fact that time in the meson frame, which is moving at .9978, is 11.74 times less than the elapsed time in the lab frame. Am I correct?

I tried to solve this by figuring out what the space time interval was between the creation of the mesons and the moment of their first half life. I took 5.4² - 5.38² (5.4 times .9978)²) = interval ². interval² = .22. Interval (or time elapsed in the meson frame) = .46 meters of time which is 11.74 times less than 5.4 meters of time in the lab frame. Is this correct?

I think this also could've been solved by looking up the gamma factor of .9978c. The reason I solved it this way instead is because the book hasn't yet gotten into this yet. They teach how to do it based on the equation for the space time interval. Could these two methods of solving the problem be compared to figuring out the side of a right triangle using the Pythagorean theorem, where we need to know the other two sides' lengths, vs. figuring out the side of a right triangle using trigonometry where we need to only know the length of one other side and the angle (where angle is comparable to speed in a special relativity problem)?

 

[Andrew Mason]

I think this also could've been solved by looking up the gamma factor of .9978c. The reason I solved it this way instead is because the book hasn't yet gotten into this yet. They teach how to do it based on the equation for the space time interval. Could these two methods of solving the problem be compared to figuring out the side of a right triangle using the Pythagorean theorem, where we need to know the other two sides' lengths, vs. figuring out the side of a right triangle using trigonometry where we need to only know the length of one other side and the angle (where angle is comparable to speed in a special relativity problem)?

This is a time dilation problem. To work it out you have apply the Lorentz transformations. The relationship between t the time between events that occur at the same place in one frame (the meson frame in this case) - called the proper time - and t' the time between same events as measured in a frame in which those events occur at different locations is: $t' = \gamma t$ where $\gamma = \frac {1}{\sqrt {1- v^2/c^2}}$ and v is the relative speed of the frames.

I get $\gamma = 15.1$. so the meson should travel 5.4x 15.1 = 81.4 m in one meson half life.

AM

 

[Ibix]

Is this correct?

No. You know the interval between emission and "half life event", because this is c times the elapsed time for the particle (and that does not agree with your 0.22 value for its square). In the lab frame you can write down $\Delta t$ in terms of $\Delta x$ and the velocity of the particles. That will allow you to determine the distance traveled by the particles in the lab frame.

The answer should, indeed, be the Lorentz gamma factor (edit - not quite - see next post). I agree Andrew Mason's answer for that.

Could these two methods of solving the problem be compared to figuring out the side of a right triangle using the Pythagorean theorem, where we need to know the other two sides' lengths, vs. figuring out the side of a right triangle using trigonometry where we need to only know the length of one other side and the angle (where angle is comparable to speed in a special relativity problem)?

More or less. The angle isn't the speed, it's the rapidity, which is $\tanh^{-1}(v/c)$. So it's a function of the speed, and has the nice property that it adds linearly, unlike velocities in relativity.

 

[Ibix]

The answer should, indeed, be the Lorentz gamma factor.

This isn't quite correct - apologies. I lost track of an assumption.

The velocity you assume in part (a) is c, whereas you are given a specified velocity, $v$, in (b). Thus the ratio should be $\gamma v/c$, not $\gamma$. I still agree Andrew Mason's value for $\gamma$.

 

[Andrew Mason]

You are correct, of course, regarding the distance compared to the distance traveled by a photon (although the difference is hardly discernible with these significant figures). But the time dilation is still: $t' = \gamma t$. So it goes farther than one would expect it to travel without time dilation by a factor of $\gamma$.

AM

 

[Ibix]

Part (a) asks you to calculate an upper bound on the possible distance by asking how far something could travel if it went at c but there were no time dilation. Part (b) asks for a full-on relativistic calculation with a specified v and the ratio of this to the result of (a), which is $\gamma v/c$. Then in the last clause it interprets this as the time dilation factor, which is of course $\gamma$.

So the question is ambiguous about what it wants, on a close reading. And slightly silly anyway because if you are assuming no time dilation why would you assume c was a limit? It would be better to have introduced the 0.9978c velocity in part (a), which would have made it a consistent Newtonian approach.

 

[NoahsArk]

Thank you for the responses.

When I redid the problem I get around .3579 for the interval which is 15 times less that the lab's time.

I was also confused a bit by part a) when it asks to assume near light speed with no time dilation.

 

[Andrew Mason]

Thank you for the responses.

When I redid the problem I get around .3579 for the interval which is 15 times less that the lab's time.

??

I was also confused a bit by part a) when it asks to assume near light speed with no time dilation.

You are testing the theory! If there was no time dilation, how far would a meson travel in one half life? If it travels 15 times farther than that, you have found evidence tending to confirm the theory.

AM

 

[NoahsArk]

I am labeling the lab frame as the frame that the mesons are moving with respect to, and that the mesons are in their separate frame.

From the lab perspective, 5.4 meters of light travel time elapse between birth of mesons and their first half life. Square that to get 29.16. Also, from the lab perspective, the mesons in the accelerator traveled a distance of 5.39 (.9978 times 5.4). Square that and you get 29.03. The difference between those two squares is .13, and the square root of .13 is .36. I am coming up with the same answer by using the spacetime interval instead of using γ. Is this not correct?

You are testing the theory! If there was no time dilation, how far would a meson travel in one half life? If it travels 15 times farther than that, you have found evidence tending to confirm the theory.

Which theory? SR? If there were no time dilation, then the mesons would travel a maximum of 5.4 meters before their half life. It would not matter how fast they were going. With time dilation, they could go infinitely far from the lab's perspective.

 

[Andrew Mason]

I am labeling the lab frame as the frame that the mesons are moving with respect to, and that the mesons are in their separate frame.

From the lab perspective, 5.4 meters of light travel time elapse between birth of mesons and their first half life. Square that to get 29.16. Also, from the lab perspective, the mesons in the accelerator traveled a distance of 5.39 (.9978 times 5.4). Square that and you get 29.03. The difference between those two squares is .13, and the square root of .13 is .36. I am coming up with the same answer by using the spacetime interval instead of using γ. Is this not correct?

You will have to explain why you are doing that. You are not applying the Lorentz transformations, which is what you have to do. You seem to be applying Pythagoras to some triangle. I am not sure why you are doing that but it is not correct.

Which theory? SR? If there were no time dilation, then the mesons would travel a maximum of 5.4 meters before their half life. It would not matter how fast they were going. With time dilation, they could go infinitely far from the lab's perspective.

Yes, SR! If there was no time dilation, the distance covered in both the lab frame and the meson frame would be the same: d = vt. I am not sure why you think it would not matter how fast they were going. I am not sure why you think they would go infinitely far with time dilation from the lab's perspective. Did you mean to say that there is no limit as to how far they would travel as c-v approached 0? If so, that would be correct.

AM

 

[Ibix]

When I redid the problem I get around .3579 for the interval which is 15 times less that the lab's time.

This isn't correct. The interval is easiest to calculate in the rest frame of the particles, where $\Delta s^2=c^2\Delta t^2-\Delta x^2$ simplifies to $\Delta s^2=c^2\Delta t^2$ because the particle is not moving. I believe you already calculated this quantity correctly as (5.4m)², and I have no idea where you would get the .3579 from.

What you then do is use that $\Delta s^2$ is invariant and that you know the relationship between $\Delta x$ and $\Delta t$ in the lab frame in order to derive the elapsed time or traversed distance in this frame.

 

[Andrew Mason]

After seeing Ibix's post, it appears that you are trying to use the space-time interval to determine the answer as a short-cut. Conceptually, one can think of a right triangle, the hypotenuse of which is ΔS which works if ΔS² > 0 (which is not necessarily the case). But this does not provide a short-cut to avoid the Lorentz transformations. In order to solve it you need to know the length of two sides. One is the hypotenuse, c²t², which you know because it is invariant. To determine the length of one of the other sides, you have to apply the Lorentz transformations.

In this case, the space-time interval between the two events (creation of meson and end of first half life) is ΔS² = c²t², since x=0 in the meson frame (see Ibix's last post). Since the interval is invariant: c²t² = ΔS² = c²t'² - Δx'². So Δx'² = c²t'² - c²t². Applying the Lorentz transformation $t' = \gamma t$ results in:

$\Delta x'^2 = c^2(\gamma^2-1)t^2 = c^2(15.1^2-1)(1.8*10^{-8})^2 = 9*10^{16} * (736)*10^{-16}= 6624$
So: $\Delta x' = \sqrt{6624} = 81.4 m.$

This is how far the meson travels in the lab frame during one half-life.

AM

 

[NoahsArk]

If I understand correctly, if we want to find the proper time that elapsed between two events, we can do that if we know the time that elapsed in some other frame (in this case the lab frame) as well as the distance between those two events in the other (lab) frame. Once we know the proper time that elapsed we can get a ratio between the time in one frame and the time in the other frame (or γ in other words). Also, I am assuming that to get γ it doesn't really matter which two events we use, as long we know the elapsed time in one frame and the distance traveled.

In this case, we know that in 5.4 light meters of lab time the mesons will have traveled a distance of 5.38 meters. The difference of those two squares is .13. .13 = the interval² which = .36 (which is close to the .3579 number that I wrote in my earlier post). So when 5.4 light meters of time passed in the lab, .36 meters of time passed in the meson frame. 5.4 is 15 times more than .36, and 5.4 times 15 = 81 meters.

I get γ=15.1γ=15.1\gamma = 15.1 . so the meson should travel 5.4x 15.1 = 81.4 m in one meson half life.

I also come out with a distance of 81 meters. I guess that means I either did it the wrong way and got lucky, or did it an alternative way which also works. I see in your most recent post though you got 76.1 meters. I haven't yet really learned the Lorentz Transformations so I will have to review the steps you took to get there.

I have no idea where you would get the .3579 from

.36 (which I approximated from .3579) is the square root of the difference between 5.4² and 5.38812²

Also, the two events that I used to calculate distance in the lab frame were not really the birth and half life of the mesons. The first event was the birth of the mesons but the second event was when the mesons traveled a distance of 5.38812 meters.

Did you mean to say that there is no limit as to how far they would travel as c-v approached 0? If so, that would be correct.

Yes that sounds more accurate then the way I stated it.

 

[Ibix]

If I understand correctly, if we want to find the proper time that elapsed between two events, we can do that if we know the time that elapsed in some other frame (in this case the lab frame) as well as the distance between those two events in the other (lab) frame. Once we know the proper time that elapsed we can get a ratio between the time in one frame and the time in the other frame (or γ in other words). Also, I am assuming that to get γ it doesn't really matter which two events we use, as long we know the elapsed time in one frame and the distance traveled.

In this case, we know that in 5.4 light meters of lab time the mesons will have traveled a distance of 5.38 meters. The difference of those two squares is .13. .13 = the interval2 which = .36 (which is close to the .3579 number that I wrote in my earlier post). So when 5.4 light meters of time passed in the lab, .36 meters of time passed in the meson frame. 5.4 is 15 times more than .36, and 5.4 times 15 = 81 meters.

I see. It's more complex than it needs to be, but yes it is correct.

 

[Andrew Mason]

If I understand correctly, if we want to find the proper time that elapsed between two events, we can do that if we know the time that elapsed in some other frame (in this case the lab frame) as well as the distance between those two events in the other (lab) frame. Once we know the proper time that elapsed we can get a ratio between the time in one frame and the time in the other frame (or γ in other words). Also, I am assuming that to get γ it doesn't really matter which two events we use, as long we know the elapsed time in one frame and the distance traveled.

In this case, we know that in 5.4 light meters of lab time the mesons will have traveled a distance of 5.38 meters. The difference of those two squares is .13. .13 = the interval2 which = .36 (which is close to the .3579 number that I wrote in my earlier post). So when 5.4 light meters of time passed in the lab, .36 meters of time passed in the meson frame. 5.4 is 15 times more than .36, and 5.4 times 15 = 81 meters.

For time-like events, if you know the distance Δx' between the events AND the time interval, t', between those events in a frame of reference, you can determine the space-time interval and, therefore, the proper time interval between those events (i.e. the time interval between events in the frame in which the events occur at the same location). That is because the proper time interval squared IS the space-time interval.

The proper time is the time between events in the frame moving at relative speed v = Δx'/t'. The proper time would be given by the space-time interval: $\Delta S^2 = (ct)^2 = (ct')^2 - \Delta x'^2$. [Note: If t is in light-m, c = 1 (i.e. 1 m per light-m) so we can dispense with the c.]

So, with t' = 5.4 light-m. and Δx' = 5.388 m., $(t)^2 = (5.4)^2 - (5.388)^2 = .1277$ so t = .3574 (light-m). Now from that, you can calculate $\gamma = t'/t = 5.4/.3574 = 15.1.$. Similarly, if t = 5.4 light-m (which is 18 nanoseconds), you can use $\gamma$ to determine t' and from that Δx':
$Δx'^2 = (t')^2 - (t)^2 = (\gamma^2-1)t^2)= (15.1^2-1)(5.4) = 81.4m.$

HOWEVER, if the other frame is NOT the frame in which the events occur at the same location, there would be two unknowns: t and Δx. In that case, the space-time interval is not sufficient to solve for these. You must use the Lorentz transformations.

AM

 

[NoahsArk]

Thank you Ibix and A.M. for your help with this. I appreciate the feedback.

 

[NoahsArk]

In chapter 3 of Spacetime Physics they give another example that sort of throws a dent in my understanding of this post's mesons problem. The general idea of this posts problem is that from the lab frame's perspective, the meson frame's clocks tick slower than lab frame's clocks, and they tick 15 times slower. Therefore, the mesons can travel 15 times farther before experiencing a half life then they would have been able to travel without time dilation.

However, the above reasoning assumes that the half life for mesons in both frames is the same. Another problem in chapter 3 points out that half lives are different in different frames based on relative speed. So, if in this post's mesons problem, the lab frame measures the meson frame's clocks as ticking 15 times slower than it's own, but also measures that mesons have half lives that are 15 times less than mesons in it's own frame, then the traveling mesons would actually go the same distance that they would have gone had there been no time dilation.

What am I missing?

 

[PeroK]

In chapter 3 of Spacetime Physics they give another example that sort of throws a dent in my understanding of this post's mesons problem. The general idea of this posts problem is that from the lab frame's perspective, the meson frame's clocks tick slower than lab frame's clocks, and they tick 15 times slower. Therefore, the mesons can travel 15 times farther before experiencing a half life then they would have been able to travel without time dilation.

However, the above reasoning assumes that the half life for mesons in both frames is the same. Another problem in chapter 3 points out that half lives are different in different frames based on relative speed. So, if in this post's mesons problem, the lab frame measures the meson frame's clocks as ticking 15 times slower than it's own, but also measures that mesons have half lives that are 15 times less than mesons in it's own frame, then the traveling mesons would actually go the same distance that they would have gone had there been no time dilation.

What am I missing?

The half life is defined in the rest frame of the meson. In a frame where the meson is moving near the speed of light, the phenomenon of meson decay will take longer, by the appropriate gamma factor. The measured half-life in this frame is, therefore, longer.

 

[Ibix]

What am I missing?

The point of the word "relativity" in relativity theory is that there are no special frames, so the mesons' half lives as measured in their rest frames are always the same. If they weren't, you'd be able to detect motion in some absolute sense by keeping mesons in a trap and measuring their half lives.

So the mesons always see their own half life as short. But, observed from the lab frame, the mesons' time is "slowed down". So according to the lab clocks the mesons live longer than their rest half-life.

As @PeroK says, half life is defined to be the half life in the meson's rest frame. So what the lab frame is measuring is not conventionally referred to as the half life.

 

[Ibix]

By the way, here are a couple of spacetime diagrams I produced for a different thread (hence why this post says "muons" throughout). Here's the lab frame:

Time goes up the page. Muons are created at the top of the atmosphere in an unrealistically regular fashion at the red crosses, and travel to the left until they reach the Earth's surface - a the blue line, which is vertical because the surface is not moving in this frame. Now here's the scenario in the muon frame:

In this frame, the red muon trails are vertical because they are not moving, but both the Earth's surface (blue line) and the top of the atmosphere (picked out by the red Xs of the muon creation events) are steeply sloped to the right. Note how much less vertical extent there is to the red lines in this frame compared to the lab frame - this is time dilation at work. In this frame, the muons don't need to survive very long to reach the ground (or for the ground to reach them, more precisely), whereas in the lab frame they need to survive much longer.

Note that these diagrams are unrealistic in that the muons are only doing 0.8c, so the time dilation factor is about a factor of ten too small. The lower diagram was illegible if I used a more realistic velocity, and there are no qualitative differences introduced by this.

 

[NoahsArk]

The half life is defined in the rest frame of the meson.

so the mesons' half lives as measured in their rest frames are always the same

Ok thank you the problem makes more sense now. So, if a certain particle's half life is let's say 1 second, then you can put a bunch of those particles in a moving ship, and keep a bunch behind on earth, and both the Earth frame and the rocket frame will say their own particles have a half life of one second? It's just that, from the perspective of the person on earth, more half lives and more seconds will go by before one half life and one second goes by in the ship?

Ibix also thank you for the diagram. Spacetime diagrams have always been tricky for me to interpret when presented from more than one reference frame. It also seems tricky to show the path of a meson falling on a spacetime diagram since the arrow of movement needs to go up with the time axis when the mesons are really falling. If I'm not mistaken though, the gist of these two pictures is that in the first one the mesons are traveling through both space and time (where the blue line represents the ground of the earth), and in the second diagram the mesons only travel along the time axis up and down.

 

[PeroK]

Ok thank you the problem makes more sense now. So, if a certain particle's half life is let's say 1 second, then you can put a bunch of those particles in a moving ship, and keep a bunch behind on earth, and both the Earth frame and the rocket frame will say their own particles have a half life of one second? It's just that, from the perspective of the person on earth, more half lives and more seconds will go by before one half life and one second goes by in the ship?

Yes. If you use the known half-life to measure time, then that it precisely what time dilation means.

 

[Ibix]

Ibix also thank you for the diagram. Spacetime diagrams have always been tricky for me to interpret when presented from more than one reference frame.

What's happening at a given time is a horizontal slice through the diagram. What's happening at a given time in another frame is a diagonal slice through the diagram. Note the presence of the grey dotted line, which shows you the diagonal line you need to use to see the Earth frame's "now".

It also seems tricky to show the path of a meson falling on a spacetime diagram since the arrow of movement needs to go up with the time axis when the mesons are really falling.

Height above the Earth at a given time is distance to the right of the blue line. So in the Earth frame "falling" means that the arrows are tilted to the left. Their height decreases with time.

In the meson frame the mesons do not move. Instead the Earth rises up to meet them - its worldline is sloped to the right.

If I'm not mistaken though, the gist of these two pictures is that in the first one the mesons are traveling through both space and time (where the blue line represents the ground of the earth), and in the second diagram the mesons only travel along the time axis up and down.

One can debate whether "travel" is the right word in this context, but basically yes. The important feature is that the vertical extent of the arrows is shortest in their rest frame, meaning that their lifetimes are shortest as measured in this frame - all other frames measure them taking longer to fall so must see their clocks ticking slowly.

 

[nitsuj]

I see. It's more complex than it needs to be, but yes it is correct.

It is, but is also less complicated.

I liked it

 

[PeterDonis]

Thread closed for moderation.

 

[PeterDonis]

A subthread that is not really germane to the main point of this thread has been moved. Thread reopened.

 

[Mister T]

What am I missing?

The lab frame needs two clocks, one located at each event, and synchronized in the lab frame. In the muon's rest frame those clocks will not be synchronized.

 

[Andrew Mason]

The lab frame needs two clocks, one located at each event, and synchronized in the lab frame. In the muon's rest frame those clocks will not be synchronized.

However, the lab frame is not using clocks. It is measuring distance that the muon travels and, using the known proper lifetime of the muon, determines that the muon has gone farther than its proper lifetime would allow.

AM

 

[PeterDonis]

The lab frame needs two clocks, one located at each event, and synchronized in the lab frame. In the muon's rest frame those clocks will not be synchronized.

This is correct if we want to be able to assign coordinates to all the events of interest in the lab frame and the muon frame. However, in the post you were responding to, the OP's confusion was much simpler: he wasn't realizing (at the time) that what he was quoting from Taylor & Wheeler was just another way of describing the effect of time dilation on the observed half-life, in the lab frame (as opposed to the muon rest frame) of the muons.

the lab frame is not using clocks. It is measuring distance that the muon travels and, using the known proper lifetime of the muon, determines that the muon has gone farther than its proper lifetime would allow.

This is correct as a description of how the muon half-life, in the lab frame, is actually observed. However, as noted above, the error the OP was making in the post @Mister T was responding to was actually much simpler.

 

[Orodruin]

You will have to explain why you are doing that. You are not applying the Lorentz transformations, which is what you have to do. You seem to be applying Pythagoras to some triangle. I am not sure why you are doing that but it is not correct.

But this does not provide a short-cut to avoid the Lorentz transformations. In order to solve it you need to know the length of two sides. One is the hypotenuse, c2t2, which you know because it is invariant. To determine the length of one of the other sides, you have to apply the Lorentz transformations.

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.

You can also use hyperbolic trigonometry to get things done easily (as geometry of Minkowski space is hyperbolic). In this case, you would have a triangle with known hyperbolic angle ($\theta = \tanh^{-1} v$) and hypothenuse (proper time $\tau$). The sought quantity is the side opposite to the hyperbolic angle and therefore
$$
x = \tau \sinh(\theta) = \tau \sinh(\tanh^{-1}v) = \frac{\tau v}{\sqrt{1-v^2}}
$$
as expected.

 

[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.