Velocity through spacetime

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I have read that the velocity of a body through spacetime is equal to 'c'.
Can I therefore draw a spacetime velocity diagram (triangle) with sides: "v" velocity through space.
"c". ------------------ time
"c" ------------------ spacetime?

runningc
 

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  • #2
PeterDonis
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I have read that the velocity of a body through spacetime is equal to 'c'.
Have you read it in an actual textbook or peer-reviewed paper?

Can I therefore draw a spacetime velocity diagram (triangle) with sides: "v" velocity through space.
"c". ------------------ time
"c" ------------------ spacetime?
No. The "velocity through spacetime" ##c## is the length of the object's 4-velocity vector. You can draw a triangle in velocity space with this as one side, and the second side being the spatial velocity ##v##; but the third side will not be ##c##.
 
  • #3
robphy
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Here's a spacetime diagram
upload_2018-10-7_7-24-50.png

In trigonometric form,
[itex]v/c\quad\ =\tanh\theta=\frac{opp}{adj}[/itex]
[itex]\gamma\qquad \ =\cosh\theta =\frac{adj}{hyp} [/itex]
[itex]\gamma(v/c)=\sinh\theta =\frac{opp}{hyp} [/itex]

(I updated the diagram by indicating the rapidity [itex]\theta[/itex] (the Minkowski-analogue of the angle).
By the way, the rapidity of the Minkowski-right-angle is infinity.
I also added the interpretation of "tangent" as "opposite over adjacent", etc...

Also note: the tangent-line to the Minkowski-circle is Minkowski-perpendicular to the radius.)
 

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Have you read it in an actual textbook or peer-reviewed paper?



No. The "velocity through spacetime" ##c## is the length of the object's 4-velocity vector. You can draw a triangle in velocity space with this as one side, and the second side being the spatial velocity ##v##; but the third side will not be ##c##.
Thank you; I was afraid it might not be that simple. I read it in "Why does E= mc^2" by Cox and Forshaw.
RC
 
  • #5
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Here's a spacetime diagram
View attachment 231822

In trigonometric form,
[itex]v/c\quad\ =\tanh\theta[/itex]
[itex]\gamma\qquad \ =\cosh\theta[/itex]
[itex]\gamma(v/c)=\sinh\theta[/itex]
Thanks very much, although I not quite sure how much it will actually help me in what I am trying to do as the diagram has 'c' and 'v' at right angles. I am working on an idea which seems to suggest that, despite Minkowsky, the Angle is variable and cannot exceed 90 deg.
RC
 
  • #6
Ibix
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I am working on an idea which seems to suggest that, despite Minkowsky, the Angle is variable and cannot exceed 90 deg.
Do be aware of PF rules on personal theories. In summary, take it to a journal - we do not discuss them here.

The diagram @robphy posted shows a four velocity (diagonal arrow, length ##c##) with its time component (vertical line, length ##\gamma c##) and its spatial component (horizontal line, length ##\gamma v##). And ##c^2=(\gamma c)^2-(\gamma v)^2##. There are a number of other things shown on there relating to other frames, but those are the relevant bits.

The angle robphy mentions is also called the rapidity, and is a way of representing velocities and gamma factors. It does, of course, vary with velocity.
 
  • #7
robphy
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The angle robphy mentions is also called the rapidity, and is a way of representing velocities and gamma factors. It does, of course, vary with velocity.
I forgot to mention rapidity and draw it in the diagram. It's updated now. Thanks.
 
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I have read that the velocity of a body through spacetime is equal to 'c'.
Can I therefore draw a spacetime velocity diagram (triangle) with sides: "v" velocity through space.
"c". ------------------ time
"c" ------------------ spacetime?

runningc
This sounds Brian Greenish.

Cheers
 
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  • #9
PeterDonis
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This sounds Brian Greenish.
Indeed, Brian Greene has used this in at least one of his pop science books. A key red flag is that he has not used it in any actual textbook or peer-reviewed paper.
 
  • #10
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I've often seen this pop-sci factoid go something like this:

"Your speed through spacetime is always ##c##. When you're at rest, your movement is entirely through time. As your speed through space increases, your speed through time decreases so that your speed through spacetime remains ##c##."

First of all, "speed through time" is word salad, isn't it? But if we have to label something "speed through time," we should consider this equation for the squared norm of the four-velocity:

##c^2 = (\gamma c)^2 - (\gamma v)^2##

Of course, ##v## is speed ("through space"), so maybe ##c## should be "speed through time"? But if so, it doesn't decrease as ##v## increases; it's a constant. Or maybe if we want to call ##\gamma v## "speed through space" (celerity magnitude), then we can call ##\gamma c## "speed through time," but now "speed through time" actually increases as "speed through space" increases, and it is in fact always the greater of the two (because ##c > v##).

Am I missing something?

Seems like the point is to convey the idea of "time components," which is fine. But can't this be done without implying that the geometry of spacetime is Euclidean? One can at least emphasize that it's only an analogy, that it isn't quite true.
 
  • #11
Ibix
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I have to say I've started to think that "four-velocity" is a bit of a misnomer. Velocity in the intuitive sense is coordinate 3-velocity, and the 4-velocity is more like the 4-d extension of a unit direction vector than it is like an extension of 3-velocity.
 
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I have to say I've started to think that "four-velocity" is a bit of a misnomer. Velocity in the intuitive sense is coordinate 3-velocity, and the 4-velocity is more like the 4-d extension of a unit direction vector than it is like an extension of 3-velocity.
I'd point out, though, that while the 4-velocity (in natural units) is indeed always a unit vector, it's also necessarily timelike and future-pointing, whereas unit 4-vectors may be either timelike (future- or past-pointing) or spacelike.
 
  • #13
PeterDonis
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I've started to think that "four-velocity" is a bit of a misnomer. Velocity in the intuitive sense is coordinate 3-velocity, and the 4-velocity is more like the 4-d extension of a unit direction vector than it is like an extension of 3-velocity.
I would agree with this. Unfortunately misnomers are common in physics. "Relativity" itself is arguably a misnomer, as Einstein often said.
 
  • #14
Matterwave
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I've often seen this pop-sci factoid go something like this:

"Your speed through spacetime is always ##c##. When you're at rest, your movement is entirely through time. As your speed through space increases, your speed through time decreases so that your speed through spacetime remains ##c##."
I've always hated this description of things...by this logic is the "speed" of a photon through space time 0? Does my "speed through space time" change if I choose to parameterize time-like curves using something other than the proper time and stop normalizing my 4-velocities to c? What in the heck is speed through time???
 
  • #15
kith
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Of course, ##v## is speed ("through space"), so maybe ##c## should be "speed through time"?
We need to divide by ##\gamma^2## in order to have ##v^2## isolated: [tex]\frac{c^2}{\gamma^2} = c^2 - v^2.[/tex] If we rearrange this, we get the equation [tex]c^2 = \frac{c^2}{\gamma^2} + v^2[/tex] which could be written as [tex]c^2 = v_{\text{time}}^2 + v_{\text{space}}^2[/tex]
where ##0 \leq v_{\text{time}} \leq c##. So there's indeed an equation which "shows" the alleged trade-off.

Note that the equation can be derived by dividing the equation for the timelike spacetime interval ##(\Delta s)^2 = (c \Delta \tau)^2= (c\Delta t)^2 - (\Delta x)^2## by ##(\Delta t)^2##. From this, we see that [tex]v_{\text{time}} = \frac{c}{\gamma} = \frac{c \Delta \tau}{\Delta t}[/tex] which I think is what's meant by the informal notion of "the speed through time".

I also haven't seen this explicitly spelled out anywhere. The concept of the "fixed speed through spacetime" originates from Epstein's book "Relativity Visualized" which doesn't contain the formulas. It does contain some numerical examples and special spacetime diagrams which involve proper time instead of coordinate time. The book is carefully argued and he makes the status of the "fixed speed through spacetime" concept clear by calling the chapter where he introduces it "The Myth". The level of the book is elementary; I recommend it to @runningc.
 
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  • #16
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We need to divide by ##\gamma^2## in order to have ##v^2## isolated: [tex]\frac{c^2}{\gamma^2} = c^2 - v^2.[/tex] If we rearrange this, we get the equation [tex]c^2 = \frac{c^2}{\gamma^2} + v^2[/tex] which could be written as [tex]c^2 = v_{\text{time}}^2 + v_{\text{space}}^2[/tex]
where ##0 \leq v_{\text{time}} \leq c##. So there's indeed an equation which "shows" the alleged trade-off.

Note that the equation can be derived by dividing the equation for the timelike spacetime interval ##(\Delta s)^2 = (c \Delta \tau)^2= (c\Delta t)^2 - (\Delta x)^2## by ##(\Delta t)^2##. From this, we see that [tex]v_{\text{time}} = \frac{c}{\gamma} = \frac{c \Delta \tau}{\Delta t}[/tex] which I think is what's meant by the informal notion of "the speed through time".

I see!

Thanks for showing that. I think it's a bizarre and unhelpful way to conceptualize the relationship between ##\gamma## and ##\beta## (it treats the frame-dependent ##\Delta x## and the Lorentz-invariant ##c \Delta \tau## as a "pair"), but now at least I understand what the heck "speed through time" is supposed to mean.
 
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  • #17
Ibix
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I'd point out, though, that while the 4-velocity (in natural units) is indeed always a unit vector, it's also necessarily timelike and future-pointing, whereas unit 4-vectors may be either timelike (future- or past-pointing) or spacelike.
Yes, although this is partly convention. There's nothing stopping you defining ##\tau'=-\tau## and hence defining a past-directed four-velocity. And there's an analogous concept for spacelike worldlines using proper distance instead of proper time - it's just less physically interesting because nothing travels on them and we wouldn't call it a velocity.
I would agree with this. Unfortunately misnomers are common in physics. "Relativity" itself is arguably a misnomer, as Einstein often said.
There's an xkcd cartoon about a guy who invents a time machine and uses it to tell Franklin to label his electrical charges the other way around. I think we need a list...
 
  • #18
martinbn
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I don't know if this is one reason, but in geometry the tangent vector to a curve is often called the velocity of the curve and its length the speed. Now, worldlines are curves and a convenient parametrization for timelike worldlines is one with tangent vector having length ##c##. So, it makes sense to say that every observer has speed (in the above sense) ##c## in space-time. This will only bring confusion in popular texts, where the typical reader wouldn't have any geometry background.
 
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  • #19
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I've often seen this pop-sci factoid go something like this:

"Your speed through spacetime is always ##c##. When you're at rest, your movement is entirely through time. As your speed through space increases, your speed through time decreases so that your speed through spacetime remains ##c##."

First of all, "speed through time" is word salad, isn't it? But if we have to label something "speed through time," we should consider this equation for the squared norm of the four-velocity:

##c^2 = (\gamma c)^2 - (\gamma v)^2##

Of course, ##v## is speed ("through space"), so maybe ##c## should be "speed through time"? But if so, it doesn't decrease as ##v## increases; it's a constant. Or maybe if we want to call ##\gamma v## "speed through space" (celerity magnitude), then we can call ##\gamma c## "speed through time," but now "speed through time" actually increases as "speed through space" increases, and it is in fact always the greater of the two (because ##c > v##).

Am I missing something?

Seems like the point is to convey the idea of "time components," which is fine. But can't this be done without implying that the geometry of spacetime is Euclidean? One can at least emphasize that it's only an analogy, that it isn't quite true.
Sorry, but I can't see what significance the equation has:
(a). It simplifies to: c^2= c^2 (b). It only has one variable. i.e. 'v'.
 
  • #20
Ibix
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Sorry, but I can't see what significance the equation has:
(a). It simplifies to: c^2= c^2 (b). It only has one variable. i.e. 'v'.
The right hand side is an explicit statement of the modulus-squared of the four-velocity. This is indeed ##c^2## whatever the value of ##v##.
 
  • #21
robphy
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Sorry, but I can't see what significance the equation has:
(a). It simplifies to: c^2= c^2 (b). It only has one variable. i.e. 'v'.
[itex]c^2 = (\gamma c)^2 - (\gamma v)^2[/itex]
is essentially the statement that
[itex]c^2 = (\cosh^2\theta) c^2 - (\sinh^2\theta) c^2[/itex] using the rapidity (the Minkowski-angle) [itex]\theta[/itex],
which uses the identity [itex]1=(\cosh\theta)^2 - (\sinh\theta)^2[/itex], effectively identifying the hyperbolas as the "circle" in Minkowski spacetime/Special Relativity.

The Euclidean analogue of the above is
[itex]1=(\cos\theta)^2 + (\sin\theta)^2[/itex], where [itex]\theta[/itex] here is the ordinary angle.
This shows up in introductory physics when we express the magnitude of a vector in terms of its x- and y-components.
If we express this in terms of slope [itex]v[/itex] instead of angle [itex]\theta[/itex],
we would write
[itex]1=(\frac{1}{\sqrt{1+v^2}})^2+ (\frac{v}{\sqrt{1+v^2}})^2[/itex]

So, in special relativity, we have the analogue
[itex]1=(\frac{1}{\sqrt{1-v^2}})^2- (\frac{v}{\sqrt{1-v^2}})^2[/itex]
associated with expressing the magnitude of a spacetime-vector in terms of its t- and x-components.
 
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  • #22
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We need to divide by ##\gamma^2## in order to have ##v^2## isolated: [tex]\frac{c^2}{\gamma^2} = c^2 - v^2.[/tex] If we rearrange this, we get the equation [tex]c^2 = \frac{c^2}{\gamma^2} + v^2[/tex] which could be written as [tex]c^2 = v_{\text{time}}^2 + v_{\text{space}}^2[/tex]
where ##0 \leq v_{\text{time}} \leq c##. So there's indeed an equation which "shows" the alleged trade-off.

Note that the equation can be derived by dividing the equation for the timelike spacetime interval ##(\Delta s)^2 = (c \Delta \tau)^2= (c\Delta t)^2 - (\Delta x)^2## by ##(\Delta t)^2##. From this, we see that [tex]v_{\text{time}} = \frac{c}{\gamma} = \frac{c \Delta \tau}{\Delta t}[/tex] which I think is what's meant by the informal notion of "the speed through time".

I also haven't seen this explicitly spelled out anywhere. The concept of the "fixed speed through spacetime" originates from Epstein's book "Relativity Visualized" which doesn't contain the formulas. It does contain some numerical examples and special spacetime diagrams which involve proper time instead of coordinate time. The book is carefully argued and he makes the status of the "fixed speed through spacetime" concept clear by calling the chapter where he introduces it "The Myth". The level of the book is elementary; I recommend it to @runningc.
How much can you really expect from physicists who write popsci books? I mean, the “ball on a trampoline” analogy gives a Kepler law with the exponents reversed. None of these popsci analogies are going to satisfy physicists, and rightfully so, as corners tend to be cut in hopes of entertaining the reader without beating them over the head with rigorous physics.

What I’m saying is, I understand why they do it, and I don’t hold that against them. I have a lot of respect for Brian Greene. He and his popsci are why I’m even a physics major. So give me the non-rigorous but cool sounding stuff in popsci books. Because those things are delicious fish bait to students deciding their majors (sadly, they are also bait to cranks, but no cracked eggs no omelets).
 
  • #23
kith
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None of these popsci analogies are going to satisfy physicists, and rightfully so, as corners tend to be cut in hopes of entertaining the reader without beating them over the head with rigorous physics.
There are two kinds of simplified explanations: the ones which can't be made rigorous and the ones which can. The rubber sheet analogy is an example of the first category. As my last post suggests, the concept of the fixed speed through spacetime is an example of the second category.

I think this is an important distinction because explanations of the first category are wrong and have to be unlearned when advancing through physics while explanations of the second category at worst present an unusual but correct viewpoint.

Of course, having a simplified explanation which can be made rigorous doesn't neccessarily mean that it is a good perspective to take if you want to understand the real theory. As @SiennaTheGr8 has remarked, it is strange to use a description of spacetime where observer-dependent coordinates for space are combined with the observer-independent proper time instead of the observer-dependent coordinate time. This doesn't reflect the way physicists work with the theory.

But it enables a laymen to build a coherent simplified view of some aspects of relativity. And for someone with a moderate background in the real physics, it gives an unusual perspective which may enrich the thinking and prompt to ask different questions.
 
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  • #24
PeterDonis
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There are two kinds of simplified explanations: the ones which can't be made rigorous and the ones which can. The rubber sheet analogy is an example of the first category.
No, it isn't. You can make it rigorous by specifying that the geometry of the rubber sheet is the geometry of a spacelike 3-surface of constant Schwarzschild coordinate time.

the concept of the fixed speed through spacetime is an example of the second category.
I agree, but I would suggest a different distinction for consideration: the distinction between simplified explanations that inevitably lead to misunderstandings and have to be unlearned, and simplified explanations that don't usually lead to misunderstandings and don't have to be unlearned (though of course their limitations will need to be recognized).

I would put both the rubber sheet and the fixed speed through spacetime in the first of my two categories. Example of the second category might be the "Einstein's elevator" explanation of the equivalence principle ("you can't tell whether the elevator is in flat or curved spacetime just from measurements inside the elevator") or Feynman's explanations of how QED works in his "QED" popular book.
 
  • #25
PeterDonis
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for someone with a moderate background in the real physics, it gives an unusual perspective which may enrich the thinking and prompt to ask different questions.
Has this ever actually happened? I have never seen the "combine observer-dependent space coordinates with proper time" method used in any actual textbook or peer-reviewed paper; the only place I've ever seen it is in pop science books. So I don't think it's actually contributed to any real physics. It's just a pet analogy of certain physicists when talking to lay people, that IMO does more harm than good.
 

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