Speed/Aging Theory: Is a 4D Universe Moving at c?

[breedb]

I have a theory and I was wondering if there is either a chance it is true, or if the theory has already been proposed:

(Assume that the universe is made up of 4 dimensions for this theory)
(Length-x, Width-y, Height-z, Time-t)

I think it makes sense that the universe is constantly moving through the time dimension at the speed of light. The reason that time passes slower, when traveling at high speeds, is that the act of traveling at the high speed skews the direction one travels.


Take for example a spacecraft that is stationary. That is to say that it is not moving in the x, y, and z, dimensions. This means according to my theory it is moving at the speed of light in the t (time) dimension.

Now suppose it begins moving at 3c/5 (three-fifths the speed of light). According to the law of the conservation of momentum, it must still be moving at c (speed of light).

So if the magnitude of the space crafts speed is still c but its moving at 3c/5 in the x dimension, and 0 in the y and z dimensions, there must be a t component of its motion that causes the sum of the x and t vector components to equal c.

Therefore Pythagorean Theorem can be used.

x^2+t^2=c^2

Where x is the speed in the x direction, t is the speed in the t direction and c is the resultant speed (speed of light). So substituting 3c/5 for x, we can find that t is 4c/5.

According to my theory this means that now the craft is only moving four-fifths the speed of light in the t direction (through time) meaning the spacecraft only ages at 4/5 the normal aging rate. This theory holds true with the notion that increased speeds will slow aging and also that a max speed is the speed of light.

Could someone please tell me if this theory either makes sense or if its already been theorized. Thank You.

 

[A.T.]

Therefore Pythagorean Theorem can be used.

x^2+t^2=c^2

Could someone please tell me if this theory either makes sense or if its already been theorized. Thank You.

It is not really a new theory, but rather an alternative geometrical interpretation of the relativity theory. Here is an interactive diagram showing the above relationship.

The topic has been discussed here:
https://www.physicsforums.com/showthread.php?t=180176
https://www.physicsforums.com/showthread.php?t=151075

 

[breedb]

Thank you. I hate it when I think I come up with something, but it turns out its been already some up with.

 

[Dale]

I hate it when I think I come up with something, but it turns out its been already some up with.

Yeah, with one sentence you summed up my years in grad school!

 

[duordi]

If time can only go forward in the time deminsion, then that would seem to indicate we are falling into a black hole in the time demension.

At least that is the only way I know of where you can not go backwards.

Has this also been suggested?

 

[Jorrie]

Take for example a spacecraft that is stationary.

Stationary relative to what?

One of the problems with the view you propose is that it assumes a preferred inertial frame of reference in which time passes 'faster' than in all other inertial frames of reference. A 'logical' choice for such a frame would have been a frame at rest relative to the cosmic microwave background (cmb) radiation (the "universe at large"). This has however been ruled out by observations. There is no frame known to us where time passes 'faster' than in all other frames of reference.

By "time passes faster" is meant that the observers in that inertial frame records a longer time interval between two events than observers in any any other inertial frame of reference.

 

[country boy]

...(Assume that the universe is made up of 4 dimensions for this theory)
(Length-x, Width-y, Height-z, Time-t)
.
.
.
Therefore Pythagorean Theorem can be used.

x^2+t^2=c^2

...

Even though your "theory" is not new, there is some insight to be gained from thinking along these lines. Your idea and the "interactive diagram" referred to in the reply from A.T. are both based on the fundamental relation

dx^2 + ds^2 = (cdt)^2

where ds is the proper "distance" equal to the proper time interval times c. The two terms on the left are the two axes in the interactive diagram and cdt is the distance along the diagonal path of the traveling object. The proper distance then behaves like an additional spatial dimension. ds has the important property that it is unchanged by transformations among inertial systems (note that this is also a propery of dy and dz when the motion is along x).

The velocities you talk about are then related by

(dx/dt)^2 + (ds/dt)^2 = c^2.

dx/dt is the observed velocity along x, while ds/dt, depicted as the vertical motion in the interactive diagram, is an unobserved velocity component.

It seems that what you call the "speed in the t direction" is actually ds/dt and that the total speed c is actually in the t direction. Do you agree? If so, it may help in relating your proposed interpretation to the interactive diagram.

 

[Bork]

Your equation can't be right, because you are saying (units of distance)^2+(units of time)^2=(units of velocity)^2, and in physics these three units should always match up in such an equation. However, Country Boy makes an excellent point that your idea is actually close to what happens in special relativity, which is that a "spacetime distance" ds^2=-(cdt)^2+(dx)^2+(dy)^2+(dz)^2 is found to be the same for all observers regardless of velocity. So if you change how fast an observer moves, their rate of time flow changes according to this relationship.

 

[breedb]

The velocities you talk about are then related by

(dx/dt)^2 + (ds/dt)^2 = c^2.

dx/dt is the observed velocity along x, while ds/dt, depicted as the vertical motion in the interactive diagram, is an unobserved velocity component.

It seems that what you call the "speed in the t direction" is actually ds/dt and that the total speed c is actually in the t direction. Do you agree? If so, it may help in relating your proposed interpretation to the interactive diagram.

Thank you. That actually was very helpful. Using ds/dt makes it easier to understand as well.

Your equation can't be right, because you are saying (units of distance)^2+(units of time)^2=(units of velocity)^2, and in physics these three units should always match up in such an equation. However, Country Boy makes an excellent point that your idea is actually close to what happens in special relativity, which is that a "spacetime distance" ds^2=-(cdt)^2+(dx)^2+(dy)^2+(dz)^2 is found to be the same for all observers regardless of velocity. So if you change how fast an observer moves, their rate of time flow changes according to this relationship.

All units in my equation are velocity units. "x" is velocity in x direction or observed velocity. "t" is a theoretical "velocity in the time direction". I'm not good at explaining what I mean by t here, but it's supposed to be the distance moved through time over the elapsed time that would be observed by a stationary observer.

 

[Chris Hillman]

Yeah, with one sentence you summed up my years in grad school!

I think you are being much too generous. Did you overlook the sign error in his "Pythagorean" law? (Presumably this should be spacetime interval?)

I don't think the OP made much sense, actually.

 

[A.T.]

Your equation can't be right, because you are saying (units of distance)^2+(units of time)^2=(units of velocity)^2, and in physics these three units should always match up in such an equation.

To get the units right, you can write the Pythagorean theorem like this:

dx^2 + (dt'*c)^2 = (dt*c)^2

where dt' is the elapsed proper time, and dt the elapsed coordinate time. So "c" is just a conversion factor between space and time units.

 

[Mortimer]

You may want to have a look at http://www.euclideanrelativity.com/links.htm , where you find more references to this use of $$dx^2+(cd\tau)^2=(cdt)^2$$.

 

[robphy]

You may want to have a look at http://www.euclideanrelativity.com/links.htm , where you find more references to this use of $$dx^2+(cd\tau)^2=(cdt)^2$$.

Note that this is, as you are aware, not identical with S&GR... which is why further discussion of it should go here https://www.physicsforums.com/showthread.php?t=103977 in the independent research forum.

I recall a discussion we had (which I have to track down) where that theory disagrees with Special Relativity's treatment of the Doppler effect.

 

[breedb]

Actually part of my theory is that space dimensions are equal to time dimensions meaning 299,792,458 meters = 1 second. Its like saying 12 inches = 1 foot. What i mean is every "time interval equivalent to one second" you have a choice to move 299,792,458 meters in the x, y or, z direction, or 1 second in the t direction. (or some combination that has an equivalent vector sum)

 

[Dale]

That idea is called geometric units: http://en.wikipedia.org/wiki/Geometrized_unit_system

 

[Jorrie]

... What i mean is every "time interval equivalent to one second" you have a choice to move 299,792,458 meters in the x, y or, z direction, or 1 second in the t direction. (or some combination that has an equivalent vector sum)

Your view has been extensively discussed in Lewis Carroll Epstein's popular book "Relativity Visualized". He calls it a "myth" and deeper study makes it clear that although valid under certain conditions, the idea is of limited use in relativity theory.