Einstein's June 30 1905 paper - help please with the maths

[larny]

I have read the Einstein June 30, 1905 paper "On the Electrodynamics of moving bodies" but I can't see how he did the following (the

text I've quoted below is on page 6 in my .pdf version):

" 1/2 [ T(0,0,0,t) + T(0,0,0,t + x'/(c-v) + x'(c+v)] = T[x',0,0,t + x'/(c-v)]

Hence, if x' be chosen infinitesimally small,

1/2 [1/(c-v) + 1/(c+v)] dT/dt= dT/dx' + 1/(c-v) dT/dt "

I don't see how x'/(c-v) and x'/(c+v) becomes 1/(c-v) and 1/(c+v) when x' is chosen infinitesimally small.

I can do it by differentiation, eg. dT/x' = 1/(c-v) + 1/(c+v) etc. but I feel he did it by a more fundamental method such as

f'(x) = [f(x + h) - f(x)]/h but I can't see how to use this technique in this case.

Any assistance will be appreciated.

 

[DrGreg]

See the thread Understanding Einstein's Math and in particular my post #6.

 

[larny]

See the thread Understanding Einstein's Math and in particular my post #6.

Thanks DrGreg.
The attachment shows how I did it some months ago, but I was not happy with it. It is some 40 years since I studied Physics, so I'm a bit rusty.

I notice that what I have done is similar to your method, but I used Einstein's short cuts re x' & t.

I'll now study your method in more detail as I don't fully understand it yet.

 

[JesseM]

See the thread Understanding Einstein's Math and in particular my post #6.

And you could also take a look at my post #4 on this thread where I tried to follow along with Einstein's 1905 derivation, making a lot of implicit steps explicit (the only thing I might change in that post: there was a step where I derived 1/2(1/(c-v) + 1/(c+v))*(dtau/dt') = dtau/dx' + (1/c-v)*(dtau/dt'), the very equation you are asking about, using an argument about zooming in on an infinitesimal region of the function tau(x',t') so it could be treated as a tilted plane, but it's probably simpler to just use the chain rule as explained in post #6 of the thread DrGreg linked to)

 

[larny]

Thanks JesseM, I'll read that and the other links to which it leads also.

 

[GrayGhost]

And you could also take a look at my post #4 on this thread where I tried to follow along with Einstein's 1905 derivation, making a lot of implicit steps explicit (the only thing I might change in that post: there was a step where I derived 1/2(1/(c-v) + 1/(c+v))*(dtau/dt') = dtau/dx' + (1/c-v)*(dtau/dt'), the very equation you are asking about, using an argument about zooming in on an infinitesimal region of the function tau(x',t') so it could be treated as a tilted plane, but it's probably simpler to just use the chain rule as explained in post #6 of the thread DrGreg linked to)

Yes, that's pretty much the way I saw it too. Einstein believed the relation between systems K and k was a linear one. Reducing x' only serves to reduce the duration of ray's roundtrip out and back, because x' represents the separation between the emitter and reflector. Reducing x', even if to a point, does not change the slope of the function, and the slope is the same on a linear function no matter what point is considered.

GrayGhost

 

[larny]

See the thread Understanding Einstein's Math and in particular my post #6.

I see that you used X’, Y, Z and T to denote the four co-ordinate variables of which tau is a function. And that Y and Z are zero.

But I don't see how you can do dX'/dx' because, as far as I can see, x' is a point on the X' axis.

I have drawn the attachment to illustrate my understanding of the situation and I can obtain Einstein's maths from it.

Does it look right?

Thanks for the responses.

 

[larny]

And you could also take a look at my post #4 on this thread where I tried to follow along with Einstein's 1905 derivation, making a lot of implicit steps explicit (the only thing I might change in that post: there was a step where I derived 1/2(1/(c-v) + 1/(c+v))*(dtau/dt') = dtau/dx' + (1/c-v)*(dtau/dt'), the very equation you are asking about, using an argument about zooming in on an infinitesimal region of the function tau(x',t') so it could be treated as a tilted plane, but it's probably simpler to just use the chain rule as explained in post #6 of the thread DrGreg linked to)

Thanks, I had not realized that tau is a plane. But of course, it must be as Einstein said that tau is linear.

 

[JesseM]

Thanks, I had not realized that tau is a plane. But of course, it must be as Einstein said that tau is linear.

I wasn't actually assuming that tau is a plane in my explanation, just that since we're dealing with infinitesimal rates of change, you're free to zoom in on an infinitesimally small neighborhood of the function tau, in which case the curvature becomes negligible (think of zooming in on a tiny region of a globe or other curved surface, as long as there aren't any singularities in the derivatives leading to sharp peaks, the tiny region should look pretty much plane-like). I thought of this because right before writing the equation 1/2(1/(c-v) + 1/(c+v))*(dtau/dt') = dtau/dx' + (1/c-v)*(dtau/dt'), Einstein wrote "Hence, if x' be chosen infinitesimally small".

 

[GrayGhost]

I have drawn the attachment to illustrate my understanding of the situation and I can obtain Einstein's maths from it. Does it look right?

Nice and interesting attempt, your graphic. It might help if you add the X-axis (wrt X,Y,Z,tau) and then draw in the light path as it travels in each system.

GrayGhost

 

[larny]

I wasn't actually assuming that tau is a plane in my explanation, just that since we're dealing with infinitesimal rates of change, you're free to zoom in on an infinitesimally small neighborhood of the function tau, in which case the curvature becomes negligible (think of zooming in on a tiny region of a globe or other curved surface, as long as there aren't any singularities in the derivatives leading to sharp peaks, the tiny region should look pretty much plane-like). Yes, I understand this. I thought of this because right before writing the equation 1/2(1/(c-v) + 1/(c+v))*(dtau/dt') = dtau/dx' + (1/c-v)*(dtau/dt'), Einstein wrote "Hence, if x' be chosen infinitesimally small".

On page 6 of my .pdf of Einstein's paper, he states & I quote:-
"In the first place it is clear that the equations must be linear on account of the properties of homogeneity which we attribute to space and time"
I assume that he made the "x' be chosen infinitesimally small" comment because he was deriving the deratives & presumably using the f'(x) = [f(x + h) - f(x)]/h technique.

 

[larny]

Nice and interesting attempt, your graphic. It might help if you add the X-axis (wrt X,Y,Z,tau) I don't understand this point, did you mean X'? and then draw in the light path as it travels in each system.

GrayGhost

I showed the ligh paths in orange wrt system K. The orange lines represent both the light paths & the distances.

As far as I can see, in system k, the light would start at X' = 0, reflect at X' = x' and return to X' = 0.

 

[GrayGhost]

I showed the ligh paths in orange wrt system K. The orange lines represent both the light paths & the distances.

The light path should travel thru the x,t plane. Your orange lines only depict the space traveled for each segment. However, no light path is drawn on your figure.

In system K, the outbound leg needs to go from the origin to the bottom of the vertical Tau₁ vector, then the return leg needs to start from there and return to the bottom of the Tau₂ vector.

In system k, the outbound leg needs to go from the origin to the top of the vertical Tau₁ vector, then the return leg needs to start from there and return to the top of the Tau₂ vector.

Also, the Tau-axis should begin at the origin and run thru the top of your vertical Tau₂ vector. That's the true Tau-axis for the k system, which intersects the origin. Then you may consider the Tau,t plane to depict the slope dTau/dt.

As far as I can see, in system k, the light would start at X' = 0, reflect at X' = x' and return to X' = 0.

That's correct. However, you drafted axes x,t, not X,t or x',t. It would help to draw the X-axis, show the light path has equal out vs back path lengths in system k, but not in system K. In doing so, you'll see that your Tau-axis should not perpendicular wrt the t-axis.

GrayGhost

 

[larny]

The light path should travel thru the x,t plane. Your orange lines only depict the space traveled for each segment. However, no light path is drawn on your figure. Yes, I understand this now.

In system K, the outbound leg needs to go from the origin to the bottom of the vertical Tau₁ vector, then the return leg needs to start from there and return to the bottom of the Tau₂ vector. Yes, I should have realized that.


In system k, the outbound leg needs to go from the origin to the top of the vertical Tau₁ vector, then the return leg needs to start from there and return to the top of the Tau₂ vector.

Also, the Tau-axis should begin at the origin and run thru the top of your vertical Tau₂ vector. So what do the Tau₁ & Tau₂ vectors represent? That's the true Tau-axis for the k system, which intersects the origin. I'll have to think about that one. Then you may consider the Tau,t plane to depict the slope dTau/dt.

That's correct. However, you drafted axes x,t, not X,t or x',t. See below. It would help to draw the X-axis, show the light path has equal out vs back path lengths in system k, but not in system K. In doing so, you'll see that your Tau-axis should not perpendicular wrt the t-axis.

GrayGhost

Thanks for the response.
As I understand it, the x & X' axes are co-incident. So why does it matter that I have not shown the X' axis?

My assumption was that the Tau value at any point Tau (x , t) would be perpendicular to the x, t plane & equal to the distance between the x, t & Tau planes. Is this not so?

If not, do they lie on the Tau plane?

I have not thought about the slope dTau/dt yet as I need the answers to my questions & want to be sure the diagam is right first

I have attached my amended diagram. Is this better? Does it still need amendment?

 

[GrayGhost]

larny,

I see what you've been doing now. You have a mixture of a spacetime diagram and a 2-space diagram (w/time implied), in 1 illustration. I haven't seen one drawn that way for quite some time, so it threw me. Your initial illustration presented Tau, without depicting graphically how Tau₁ = (Tau₀+Tau₂)/2.

In your latest illustration, move the commencement of your Tau-axis to the origin of system K. Because actually, the origin of the 2 systems K and k coincide at t=Tau=0 in Einstein's model, and this is typically the proper way to commence in spacetime diagrams. Leave the other end as is, intersecting the top of the vertical Tau₂ vector. Also, your vertical K-axis should instead be called "k at emission", because they are not really axes but rather pointers (which verbiage) points upon the x-axis. Your x-axis should be renamed x,X, which doubles up as 2 axes, one for each system always colinear ... where x and X do not move (in the way you had shown it) in the updated illiustration. Lastly, you could add a dashed line-of-simultaneity forsystem k at time Tau₁, which would basically be another Tau-axis (but dashed) intersecting the top of your vertical Tau₁ vector, and the X-axis.

You can then add another line that connects the top of the vertical tau₂ vector with the point now designated t₂ upon the t-axis ... and that'll form the triangle you need within the Tau,t plane to depict graphically the dTau/dt slope.

Wrt your question regarding whether Tau and t are orthogonal, they are not far as how they are oriented in the fullest scope of spacetime (of which we are not privied). They are angularly rotated wrt one another in the spacetime (as are X and x) due to the relative motion. Although it seems that x & X are always colinear, they are not in reality, just as t & T are not and for the same reason. Because we sense time differently than space, we cannot causally observe the relative direction of arrows-of-time. We can only observe the readout of the clock and the relativistic effects (eg time dilation and length contraction), and then reason by deduction that the arrows-of-time are angularly rotated.

GrayGhost

 

[GrayGhost]

Anyone,

This in relation to a post I made in another thread ghWellsJr ...

In OEMB, it says this ...

If we assume that the result proved for a polygonal line is also valid for a continuously curved line, we arrive at this result: If one of two synchronous clocks at A is moved in a closed curve with constant velocity until it returns to A, the journey lasting t seconds, then by the clock which has remained at rest the traveled clock on its arrival at A will be ½tv²/c² second slow.

He begins with the time transformation ...

Tau = [1/(1-v²/c²)^1/2](t-vx/c²)​

and after substituting vt for x, he gets this ...

Tau = t(1-v²/c²)^1/2 = t - [1-(1-v²/c²)^1/2]t​

and he then says that the accelerated clock must lag the inertial clock on return by ...

Tau = ½tv²/c² ... up to magnitudes of fourth and higher order​

OK, so all's good. However, let's consider a constant relative velocity of v=0.866, where gamma = 2. The accelerating clock should always run at a factor of 1/gamma slower per the always-inertial clock, including over the interval on the whole. Well, 1/gamma = 1/2 = 50%. So if t = 10 sec over the interval, then Tau = t/gamma = 10/2 = 5 sec over the same interval ...

So the LAG = 5 sec upon return, which requires the accelerated clock read Tau₀ + 5 sec when the inertial clock reads t₀ + 10 sec.

Now consider Einstein's soln (up to 4th order magnitudes), where Einstein said the accelerated clock (upon return) lags the always inertial clock by ½tv²/c². We plug in v=0.866c, and the lag is ...

Lag = ½tv²/c²
Lag = ½(10s)(0.866)²/(1)²
Lag = (5s)(0.866)²
Lag = 3.74978 sec

So the Lag = 3.74978 sec upon return, which requires the accelerated clock read Tau₀ + 3.73978 sec when the inertial clock reads t₀ + 10 sec.

The lag should be 5 sec. So is the difference here due to limiting the calculation to a 4th order magnitude?

GrayGhost

 

[JesseM]

So the Lag = 3.74978 sec upon return, which requires the accelerated clock read Tau₀ + 3.73978 sec when the inertial clock reads t₀ + 10 sec.

The lag should be 5 sec. So is the difference here due to limiting the calculation to a 4th order magnitude?

GrayGhost

Yes, he's just doing a Taylor series for the function f(v) = (1 - (1 - v²/c²)^1/2), expanded around v=0. And he's only taking the first three terms, i.e.

f(0) + f'(0)*(v-0) + (1/2)*f''(0)*(v-0)²

so with f'(v) = (1 - v²/c²)^-1/2 * v/c²
and
f''(v) = -(1/2)*(1 - v²/c²)^-3/2 * v/c² + (1 - v²/c²)^-1/2 * (1/c²)
we have f(0)=0, f'(0)=0, and f''(0)=1/c². Plug that into the first three terms of the Taylor series and you get (1/2)*v²/c², the term he multiplies by t to get the time lag. If you included more terms, it would more closely agree with (1 - (1 - v²/c²)^1/2).

 

[GrayGhost]

JesseM,

Ahhh, so the Taylor series is also the Maclaurin series. I like the Wiki animation, which presents the solns more accurate with added terms in the series. Thank you much ! Highly appreciated. It's been so long since I looked at the Maclaurin series, I can't even remember :)

GrayGhost

 

[larny]

larny,

I see what you've been doing now. You have a mixture of a spacetime diagram and a 2-space diagram (w/time implied), in 1 illustration. I haven't seen one drawn that way for quite some time, so it threw me. Your initial illustration presented Tau, without depicting graphically how Tau₁ = (Tau₀+Tau₂)/2. I can't at the moment think how I would do this. However, see my first comment below.

In your latest illustration, move the commencement of your Tau-axis to the origin of system K. Ironically, this is how I drew it initially, but changed it as I thought you must have meant (by "origin") the origin of k because that would make the Tau axis orthagonal to the X axis. Because actually, the origin of the 2 systems K and k coincide at t=Tau=0 in Einstein's model, and this is typically the proper way to commence in spacetime diagrams. Leave the other end as is, intersecting the top of the vertical Tau₂ vector. Also, your vertical K-axis should instead be called "k at emission", because they are not really axes but rather pointers (which verbiage) points upon the x-axis. Your x-axis should be renamed x,X, which doubles up as 2 axes, one for each system always colinear ... where x and X do not move (in the way you had shown it) in the updated illiustration. Lastly, you could add a dashed line-of-simultaneity forsystem k at time Tau₁, which would basically be another Tau-axis (but dashed) intersecting the top of your vertical Tau₁ vector, and the X-axis.

You can then add another line that connects the top of the vertical tau₂ vector with the point now designated t₂ upon the t-axis ... and that'll form the triangle you need within the Tau,t plane to depict graphically the dTau/dt slope.

Wrt your question regarding whether Tau and t are orthogonal, they are not far as how they are oriented in the fullest scope of spacetime (of which we are not privied). They are angularly rotated wrt one another in the spacetime (as are X and x) due to the relative motion. Although it seems that x & X are always colinear, they are not in reality, just as t & T are not and for the same reason. Because we sense time differently than space, we cannot causally observe the relative direction of arrows-of-time. We can only observe the readout of the clock and the relativistic effects (eg time dilation and length contraction), and then reason by deduction that the arrows-of-time are angularly rotated.

GrayGhost

Thanks agains GrayGhost.
I need a day or 2 to think through what you & JesseM have written so far.

However, one question.

In space, we know that z = ax + by +c is the eq of a plane.

Now, Einstein's eq for Tau can be written as Tau = gt + hx' where g & h are functions of c & v.

Then, when we "substitute x' for its value" Tau can be written as Tau = pt + q where p & qx are functions of c & v.

So my question is - if the eq for z is a plane in space, then I asume that the eqs for Tau are also planes?

Einstein seems to imply this by his comment "it is clear that the equations must be linear on account of the properties of homogeneity which we attribute to space and time"

I know we are talking about space time, but an equation is an equation, so I don't understand why the Tau "vectors" are not perpendicular to the x - t plane.

 

[GrayGhost]

In space, we know that z = ax + by + c is the eq of a plane.

Now, Einstein's eq for Tau can be written as Tau = gt + hx' where g & h are functions of c & v.

Let's see if it is ...

Tau = gamma*(t-vx/c²)
Tau = gamma*t-gamma*vx/c²

Given c is invariant and v a variable "held constant" (since motion inertial), then ...

Tau = gamma*t-gamma*vx/c² ... call gamma "g", then ...
Tau = g*t-g*vx/c²
Tau = gt-(gv/c²)x ... call gv/c² a constant "h" ...
Tau = gt-hx​

Now if in terms of x', where x = x'+vt ...

Tau = gt-hx
Tau = gt-h(x'+vt)
Tau = gt-hx'-hvt
Tau = (gt-hvt)-hx'
Tau = (g-hv)t-hx' ... call g-hv a constant "G" ...
Tau = Gt-hx'​

So, I say you are right. That said, wrt your question regarding whether Tau is a plane ...

First, a plane is 2 dimensional in 3-space. Let's assume that a plane in 4-space (4d spacetime) is by anaolgy "a volume". I see Tau as a point in time, which may be considered a volume of 3-space (X,Y,Z) at that time Tau. X would be a point in space, which may be considered a volume of 3-space (Y,Z,Tau) at that location X. Y would be a point in space, which may be considered a volume of 3-space (X,Z,Tau) at that location Y. Z would be a point in space, which may be considered a volume of 3-space (X,Y,Tau) at that location Z. The intersection of all four of these 3d-volumes (the LT resolved values) should be a single point in 4-space (ie 4d spacetime ... X,Y,Z,Tau).

Then, when we "substitute x' for its value" Tau can be written as Tau = pt + q where p & qx are functions of c & v.

Indeed.

Tau = Gt-hx' ... given x' is a variable "held constant", then call -hx a constant q ...
Tau = Gt+q

So my question is - if the eq for z is a plane in space, then I asume that the eqs for Tau are also planes?

Einstein seems to imply this by his comment "it is clear that the equations must be linear on account of the properties of homogeneity which we attribute to space and time"

I know we are talking about space time, but an equation is an equation, so I don't understand why the Tau "vectors" are not perpendicular to the x - t plane.

You know, this is an interesting question actually. We assume that time t is perpendicular to x, and time Tau perpendicular to X. This much is certain. Are they? Mathematically, we set them orthogonal to each other because they are independent variables. However, should Tau be perpendicular to x or t? Should t be perpendicular to X or Tau? Well if they should, then Tau and t must be perpendicular to each other. Einstein's LT's define how points in one system are transformed into the other system, both inertial. Therefore, the 2 systems are not independent "of each other", but rather related per the LTs. The LTs are these ...

Tau = gamma*(t-vx/c²)
X = gamma*(x-vt)
Y = y
Z = z

where gamma = 1/(1-v²/c²)^1/2

So Y and y should be parallel, and Z and z should be parallel. Tau & t should not be parallel, and X & x should not be parallel. They should not be orthogonal either, because they are "always" dependent upon each other, being transformations between 2 differing systems of relative v>0. There exists an angle between the Tau and t axes and X and x axes, which depends only on the relative v.

BTW ... I don't see a problem with modeling Tau perpendicular to the x,t plane, so long as your Tau values properly relate to the other system (x and t) per the LTs. It's not the typical way of presenting a spacetime illustration though. Modeling it as such comes at an expense wrt overall meaning of the mechanism, IMO.

GrayGhost

 

[GrayGhost]

...

GrayGhost: I see what you've been doing now. You have a mixture of a spacetime diagram and a 2-space diagram (w/time implied), in 1 illustration. I haven't seen one drawn that way for quite some time, so it threw me. Your initial illustration presented Tau, without depicting graphically how Tau1 = (Tau0+Tau2)/2.

larny: I can't at the moment think how I would do this. However, see my first comment below.​

GrayGhost: In your latest illustration, move the commencement of your Tau-axis to the origin of system K.

larny: Ironically, this is how I drew it initially, but changed it as I thought you must have meant (by "origin") the origin of k because that would make the Tau axis orthagonal to the X axis.​

Your original illustration was a mix of spacetime diagram and a 2-space figure with time implied, all overlaid on a single illustration. While there's nothing wrong with it, it lacks some meaning far as the overall meaning of the theory is concerned. I was trying to figure out a way to make your figure conform more with standard spacetime diagrams. However, it cannot be done in the way that I asked. You're right, the x,X axes cannot remain colinear if you add the Tau-aaxis as I requested. On a standard Minkowski illustration, they are not colinear in fused 4-space, even though they appear as such as they move thru 3-space over time. So, it seems that the only good way to draft it is in standard Minkowski way (or similar), assuming one wishes to have as much meaning on a single page as possible.

Are you familiar with Minkowski spacetime diagrams at all?

GrayGhost

 

[larny]

Your original illustration was a mix of spacetime diagram and a 2-space figure with time implied, all overlaid on a single illustration. While there's nothing wrong with it, it lacks some meaning far as the overall meaning of the theory is concerned. I was trying to figure out a way to make your figure conform more with standard spacetime diagrams. However, it cannot be done in the way that I asked. You're right, the x,X axes cannot remain colinear if you add the Tau-aaxis as I requested. On a standard Minkowski illustration, they are not colinear in fused 4-space, even though they appear as such as they move thru 3-space over time. So, it seems that the only good way to draft it is in standard Minkowski way (or similar), assuming one wishes to have as much meaning on a single page as possible.

Are you familiar with Minkowski spacetime diagrams at all?

GrayGhost

Yes but I'm rusty on it. I studied it some years ago when I had my last burst on SR.

I have attached a scan of the title page of the book I have and a scan of my scribbles on a scan of a Minkowski spacetime diagram from page 78 of that book.

I need to revise it, but first I want to read through all of the posts so far & think them through.

I found these diagrams rather abstract thus my attemps to make graphical sense of Einstein's paper and his Tau derivation.

 

[GrayGhost]

Yes but I'm rusty on it. I studied it some years ago when I had my last burst on SR.

I have attached a scan of the title page of the book I have and a scan of my scribbles on a scan of a Minkowski spacetime diagram from page 78 of that book.

I need to revise it, but first I want to read through all of the posts so far & think them through.

I found these diagrams rather abstract thus my attemps to make graphical sense of Einstein's paper and his Tau derivation.

I would recommend you forget about trying to make sense of SR by starting with plane geometry. Your graph is fine. Your t = 2c/x should be t = x/c ... which means the photon after traveling for time t traverses a distance x = ct. So t = x/c, and thus x = ct. The light path has a 45 deg slope on Minkowski spacetime diagrams because Minkowski normalized light's speed at c=1, so 1 unit of distance per 1 unit of time ... eg 300k km/sec or 1 lt-hr/ 1 hr. The slanted time axis is the direction of said moving observer thru spacetime per the stationary observer (who owns the vertical time axis), and the slanted path cannot reach 45 deg because that's the speed of light. The moving fellow deems himself stationary, and so he perceives said motion thru spacetime as "I'm just sitting here passing thru time". If he drafts the spacetime diagram from his POV, his time axis is vertical, and the other fellow is slanted to the left (by the same slant, oppositiely though). The moving fellow holds his own x'-axis perpendicular to his own t'-axis. Per the observer moving relatively though, his t' axis is slanted (not vertical) and his spatial axis x' is slanted. It's slanted by the same angle from the light path that his t'-axis is ... because the light path must bisect all time and space axes wrt any cooridnate system, since light's speed is c = 1-unit-time per 1-unit-space = unity.

The eqn below is the invariant spacetime interval ...

s² = (ct)² - (x² + y² + z²)​

For any numeric substitutions of x,y,z,t one obtains "a distance" within a fused 4-dimensional spacetime between any 2 defined events. Thus, it's a line-element. The numerical result is also equivalent to the proper time experienced by the fellow who travels inertially between the 2 points, his own duration to transit between the 2 events. Remember, although others see him in motion traversing space over time, he assumes himself stationary and passing only thru time t' (or ct' if you prefer). So per he, s is nothing more than the proper time (duration) he experienes between one event and a subsequent event which both occur at his own location. It's an invariant, and all in the cosmos must agree, and do per the LTs.

Considering only motion along x ...

s² = (ct)² - (x² + y² + z²)

s² = (ct)² - x²

Assuming x is the location (wrt origin) of an observer at time vt, then ...

s² = (ct)² - (vt)²

s = (c²-v²)^1/2 t

s = ct(1-v²/c²)^1/2

s = ct(1/gamma)

and since s is numerically equal to t', and given c is normalized to c=1, then ...

s = ct(1/gamma)
t' = t(1/gamma)

And so the moving fellow (of x,y,z,t) is dilated by the factor gamma wrt the fellow who travels direct between both events (of x',y',z',t'). Note that folks in relative motion wrt the system x',y',z',t' cannot possibly exist AT both events.​

GrayGhost

 

[larny]

Thanks again GrayGhost,
I'll take your advice & concentrate on the Minkowski approach.