<p>I think, you are trying to describe a <a href="http://en.wikipedia.org/wiki/World_line" rel="nofollow">world line</a>.</p>

CC BY-SA 3.0
2016-10-26T18:08:44.440Z
@SirCumference Converted to wiki at Gerald's request so that others can make the necessary modifications to beef up the answer. Gerald is currently busy.
called2voyage
CC BY-SA 3.0
2014-01-11T01:16:53.187Z
Excellent answer! I really struggled in the comment section above... :) And then I can use vector calculations to calculate a lot of stuff such as Einstein shift etc I guess. I just express velocity as a function of mass and energy, derived from 0.5mv^2=E? And define everything to have velocity of c in 4D. As you know I am not a physicist, so I ask a lot... :)
frodeborli
CC BY-SA 3.0
2014-01-11T02:44:56.687Z
@frodeborli: That's not the correct relation. The correct relation is $(mc^2)^2 = E^2 - (pc)^2$, which is has the geometrical meaning of mass being the magnitude of the four-momentum vector.
Stan Liou
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2014-01-11T14:16:39.430Z
@StanLiou Magnitude? Then I am getting This world line wring, still. The magnitude would, as I see it always be c in four velocities. Then I want to apply an acceleration inversely squared by the distance to another particle, on the time dimension.
frodeborli
CC BY-SA 3.0
2014-01-11T14:25:34.150Z
I am unsure of how to apply the acceleration on the four vector yet, but I assume it'll be possible.
frodeborli
CC BY-SA 3.0
2014-01-11T14:31:17.673Z
@StanLiou I think I get it. Four-momentum is just four velocity but with mass-energy. But energy is derived from among other things, velocity in three dimensions, ignoring time.
frodeborli
CC BY-SA 3.0
2014-01-11T00:11:06.057Z
2014-01-11T00:11:06.057Z
Gerald
<p>This is a question, where I start with some assertions:</p>

<p>Try to consider the universe as a four-coordinate system, x,y,z,t where t is time and where we view a change in t as a change in position, and as a velocity in the same way as any other change in either of the other components of the coordinate. Everything is then moving at the speed of c. The difference between a photon and another particle would be that a photon "spends" its velocity in any of the x, y or z dimensions - but not in the time dimension.</p>

<p>This implies that all photons are on the same coordinate in the t-dimension - only the x,y,z values change while t remains 0. It also implies that time would appear to stop for something travelling in only the other dimensions, while it would pass fastest for something completely stationary.</p>

<p>You can then also consider space-time as a function of acceleration. A physical objects' coordinate could be derived from its velocity, which would have to be derived from its acceleration. Acceleration would be a function of entropy, or "age" I believe.</p>

<p>Is this the basis of general relativity? Is it at least an initial starting point for general relativity, or is it completely wrong?</p>

<p>Would spin have to be part of the coordinate system?</p>

CC BY-SA 3.0
2014-01-10T12:28:02.293Z
No, it is completely wrong. Everything about this is mistaken. The first mistake is confusion between four-velocity and speed, aided by ignorance of the metric. Sorry, but I don't think this is salvageable.
Stan Liou
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2014-01-10T13:19:18.457Z
@StanLiou It would be better if you could help him understand instead of just saying he's wrong
Eduardo Serra
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2014-01-10T13:25:51.010Z
@StanLiou Can't you just help me and provide one observation or calculation where this view does not match? I allows my intuition to explain Einstein shift and relative time, and why nothing can move faster than the speed of light, it even explains the twin paradox in an intuitive way.
frodeborli
CC BY-SA 3.0
2014-01-10T15:45:06.780Z
@frodeborli, I would just recommend you to take a look at some introductory book, e.g. Stephani or Schutz.
Alexey Bobrick
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2014-01-10T16:09:24.470Z
All books I've read seem to agree that it is Doppler shift due to expansion. I want to read some book or paper that explores the possibility of it being Einstein shift. Preferably one that concludes that it must be Doppler, so I can understand why everybody is dismissing cosmic Einstein shift.
frodeborli
CC BY-SA 3.0
2014-01-10T17:33:55.273Z
@frodeborli: The problems in your question start with the sentence "Everything is then moving at the speed of c." and continue further on. If you explained what you mean in terms of 4-velocities, world lines and 4-dimensional spacetime, it would perhaps be easier to understand your question. All these notions are standard and are explained in most introductory books.
Alexey Bobrick
CC BY-SA 3.0
2014-01-10T17:46:03.193Z
I am a computer programmer. Vector calculations involves sin/cos/tan. Going from 2D programming to 3D was fairly straightforward. Would involving time as a fourth dimension be correct. That is, is the magnitude of a "velocity" vector in a 4D space-time equal to c in that case?  It would allow me to get a "velocity" for t, as a function of the velocity vector of x,y and z. Also, I could get x, as a function of y,z and t etc.
frodeborli
CC BY-SA 3.0
2014-01-10T21:31:28.987Z
The correct GR terminology is four vector (http://en.wikipedia.org/wiki/Four-vector). I agree with @AlexeyBobrick - please take a look at nearly any introductory GR textbook for 4 dimensional (3 space + 1 time) coordinate systems, and any introductory cosmology textbook for basic information before dismissing work that has already been done on the subject of your question.
astromax
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2014-01-10T21:32:40.790Z
Alternatively, please edit your question to ask about the basics which you seem to be confused about.
astromax
CC BY-SA 3.0
2014-01-10T21:47:44.190Z
@EduardoSerra: I provided two things he should start with to understand this better, but I obviously can't teach someone GTR in a comment section. So what's the problem?
Stan Liou
CC BY-SA 3.0
2014-01-10T21:53:58.247Z
@fredobrli: Again, look up four-velocity and the meaning of metric, even in special relativity. You'll immediately see that the magnitude of the four-velocity for light is *zero* and that in an inertial frame, light moves as much in space as in time, directly contradicting your claims.
Stan Liou
CC BY-SA 3.0
2014-01-10T23:13:23.900Z
I've done something else. I want to figure out if einstein shift = sqr(c^2 - a^2) / sqr(c^2 - b^2). I just need to figure out a and b.
frodeborli
CC BY-SA 3.0
2014-01-10T23:27:20.500Z
Time is like a "sqrt(-1) times space"-dimension. See Minkowski space: http://en.wikipedia.org/wiki/Minkowski_space
Gerald
CC BY-SA 3.0
2014-01-11T01:40:24.103Z
Intuitively, I should get Einstein shift from sqr(c^2-sqr(2E1/m1))/sqr(c^2-sqr(2E2/m2)) E1 and m1 is energy and mass of object 1, and the same for E2 and m2?
frodeborli
CC BY-SA 3.0
2014-01-13T16:13:29.457Z
I have closed the question as too broad because the misconceptions in the question make it too difficult to ascertain the value of this Q&A. I chose "too broad" as the close reason because, as Stan Liou observes, the misconceptions are too wide ranging to correct summarily in this space.
called2voyage
CC BY-SA 3.0
2014-01-13T19:40:39.067Z
Well, the question is if the magnitude of time - in a four-velocity vector - is inversly proportional to the velocity in the three other (spatial) dimensions, so that the magnitude of the vector is always equal to the speed of light. I don't think it is too broad.
frodeborli
<general-relativity>
Age and relativity
frodeborli
# Age and relativity

(Asked by: username_0 on 2014-01-10)

This is a question, where I start with some assertions:

Try to consider the universe as a four-coordinate system, x,y,z,t where t is time and where we view a change in t as a change in position, and as a velocity in the same way as any other change in either of the other components of the coordinate. Everything is then moving at the speed of c. The difference between a photon and another particle would be that a photon "spends" its velocity in any of the x, y or z dimensions - but not in the time dimension.

This implies that all photons are on the same coordinate in the t-dimension - only the x,y,z values change while t remains 0. It also implies that time would appear to stop for something travelling in only the other dimensions, while it would pass fastest for something completely stationary.

You can then also consider space-time as a function of acceleration. A physical objects' coordinate could be derived from its velocity, which would have to be derived from its acceleration. Acceleration would be a function of entropy, or "age" I believe.

Is this the basis of general relativity? Is it at least an initial starting point for general relativity, or is it completely wrong?

Would spin have to be part of the coordinate system?

**Tags:** general-relativity
> **Comments:**
> **username_1**: No, it is completely wrong. Everything about this is mistaken. The first mistake is confusion between four-velocity and speed, aided by ignorance of the metric. Sorry, but I don't think this is salvageable.  
> **username_2**: @StanLiou It would be better if you could help him understand instead of just saying he's wrong  
> **username_0**: @StanLiou Can't you just help me and provide one observation or calculation where this view does not match? I allows my intuition to explain Einstein shift and relative time, and why nothing can move faster than the speed of light, it even explains the twin paradox in an intuitive way.  
> **username_3**: @frodeborli, I would just recommend you to take a look at some introductory book, e.g. Stephani or Schutz.  
> **username_0**: All books I've read seem to agree that it is Doppler shift due to expansion. I want to read some book or paper that explores the possibility of it being Einstein shift. Preferably one that concludes that it must be Doppler, so I can understand why everybody is dismissing cosmic Einstein shift.  
> **username_3**: @frodeborli: The problems in your question start with the sentence "Everything is then moving at the speed of c." and continue further on. If you explained what you mean in terms of 4-velocities, world lines and 4-dimensional spacetime, it would perhaps be easier to understand your question. All these notions are standard and are explained in most introductory books.  
> **username_0**: I am a computer programmer. Vector calculations involves sin/cos/tan. Going from 2D programming to 3D was fairly straightforward. Would involving time as a fourth dimension be correct. That is, is the magnitude of a "velocity" vector in a 4D space-time equal to c in that case?  It would allow me to get a "velocity" for t, as a function of the velocity vector of x,y and z. Also, I could get x, as a function of y,z and t etc.  
> **username_4**: The correct GR terminology is four vector (http://en.wikipedia.org/wiki/Four-vector). I agree with @AlexeyBobrick - please take a look at nearly any introductory GR textbook for 4 dimensional (3 space + 1 time) coordinate systems, and any introductory cosmology textbook for basic information before dismissing work that has already been done on the subject of your question.  
> **username_4**: Alternatively, please edit your question to ask about the basics which you seem to be confused about.  
> **username_1**: @EduardoSerra: I provided two things he should start with to understand this better, but I obviously can't teach someone GTR in a comment section. So what's the problem?  
> **username_1**: @fredobrli: Again, look up four-velocity and the meaning of metric, even in special relativity. You'll immediately see that the magnitude of the four-velocity for light is *zero* and that in an inertial frame, light moves as much in space as in time, directly contradicting your claims.  
> **username_0**: I've done something else. I want to figure out if einstein shift = sqr(c^2 - a^2) / sqr(c^2 - b^2). I just need to figure out a and b.  
> **username_5**: Time is like a "sqrt(-1) times space"-dimension. See Minkowski space: http://en.wikipedia.org/wiki/Minkowski_space  
> **username_0**: Intuitively, I should get Einstein shift from sqr(c^2-sqr(2E1/m1))/sqr(c^2-sqr(2E2/m2)) E1 and m1 is energy and mass of object 1, and the same for E2 and m2?  
> **username_6**: I have closed the question as too broad because the misconceptions in the question make it too difficult to ascertain the value of this Q&A. I chose "too broad" as the close reason because, as Stan Liou observes, the misconceptions are too wide ranging to correct summarily in this space.  
> **username_0**: Well, the question is if the magnitude of time - in a four-velocity vector - is inversly proportional to the velocity in the three other (spatial) dimensions, so that the magnitude of the vector is always equal to the speed of light. I don't think it is too broad.  

---

## 1 answers

### Answer by username_5 on 2014-01-11. Score: 2 | Accepted answer

I think, you are trying to describe a [world line](http://en.wikipedia.org/wiki/World_line).

> **Comments:**
> **username_0**: Excellent answer! I really struggled in the comment section above... :) And then I can use vector calculations to calculate a lot of stuff such as Einstein shift etc I guess. I just express velocity as a function of mass and energy, derived from 0.5mv^2=E? And define everything to have velocity of c in 4D. As you know I am not a physicist, so I ask a lot... :)  
> **username_1**: @frodeborli: That's not the correct relation. The correct relation is $(mc^2)^2 = E^2 - (pc)^2$, which is has the geometrical meaning of mass being the magnitude of the four-momentum vector.  
> **username_0**: @StanLiou Magnitude? Then I am getting This world line wring, still. The magnitude would, as I see it always be c in four velocities. Then I want to apply an acceleration inversely squared by the distance to another particle, on the time dimension.  
> **username_0**: I am unsure of how to apply the acceleration on the four vector yet, but I assume it'll be possible.  
> **username_0**: @StanLiou I think I get it. Four-momentum is just four velocity but with mass-energy. But energy is derived from among other things, velocity in three dimensions, ignoring time.  
> **username_6**: @SirCumference Converted to wiki at Gerald's request so that others can make the necessary modifications to beef up the answer. Gerald is currently busy.  

---

