Q&A

# Post History

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1 answer  ·  posted over 1 year ago by celtschk‭  ·  last activity over 1 year ago by dmckee‭

#3: Post edited by celtschk‭ · 2021-01-10T12:23:11Z (over 1 year ago)
• In special relativity, spacetime coordinates are typically given as
• $$(ct, x, y, z) \tag{S}$$
• with the metric being either
• $$g = \operatorname{diag}(-1,1,1,1) \tag{+}$$
• or
• $$g = \operatorname{diag}(1,-1,-1,-1) \tag{-}$$
• depending on which sign convention the author prefers (the second one seems to be most commonly used one).
• The effect is that all coordinates, including the time coordinate, are given in space units. On the other hand, *proper* time $\tau$ is always given without the factor $c$ (i.e. not as $c\tau$), that is in time units.
• Now there is no fundamental reason to use space units instead of time units, in particular for the $(-)$ case in which *timelike* intervals get positive squared metric. This would mean to specify points as
• $$(t,x/c,y/c,z/c) \tag{T}$$
• Now it seems clear to me why this option is not used: With the standard convention the 4-equivalents of 3-vectors get the same units as the corresponding 3-vectors. For example, the 4-momentum vector
• $$p=(mc \frac{\mathrm dt}{\mathrm d\tau}, • m\frac{\mathrm dx}{\mathrm d\tau}, • m\frac{\mathrm dy}{\mathrm d\tau}, • m\frac{\mathrm dz}{\mathrm d\tau})$$
• has actually the units of a momentum; in particular the space components directly give the 3-momentum.
• However I think there is a better way to do this. This is based on the observation that for coordinate systems like polar coordinates not all coordinates have the same unit, and the unit difference is then reflected in the metric. Therefore I see no reason not to do the same also for spacetime coordinates.
• In this scheme, the spacetime coordinates would be written simply as
• $$(t,x,y,z) \tag{M}$$
• and the corresponding metric (choosing the length to be in spatial units, like in the standard convention) would be
• $$g = \operatorname{diag}(-c^2,1,1,1) \tag{+'}$$
• or
• $$g = \operatorname{diag}(c^2,-1,-1,-1) \tag{-'}$$
• This convention would IMHO have several advantages:
• * Unlike the standard convention $(S)$, $(M)$ would be consistent with the fact that *proper* time is always given without the $c$ factor in the context of relativity.
• * But unlike $(T)$ (and unlike the option to just use $c\tau$ for the proper time for consistency), the choice $(M)$ keeps the usual units for the spacelike components of 4-vectors. For example, the four-momentum in that convention is
• $$p=(m\frac{\mathrm dt}{\mathrm d\tau}, • m\frac{\mathrm dx}{\mathrm d\tau}, • m\frac{\mathrm dy}{\mathrm d\tau}, • m\frac{\mathrm dz}{\mathrm d\tau})$$
• Clearly the space components are the 3-momentum in usual units. The time component (which gives the energy) isn't in conventional units (instead it is in mass units, thus giving the “relativistic mass” $E/c^2$), but it isn't in the standard convention either (where the time component is $E/c$).
• Now my question is: Is there any downside (other than “it's not what is commonly done”) to the convention $(M)$?
• In special relativity, spacetime coordinates are normally given as
• $$(ct, x, y, z) \tag{S}$$
• with the metric being either
• $$g = \operatorname{diag}(-1,1,1,1) \tag{+}$$
• or
• $$g = \operatorname{diag}(1,-1,-1,-1) \tag{-}$$
• depending on which sign convention the author prefers (the second one seems to be most commonly used one).
• The effect is that all coordinates, including the time coordinate, are given in space units. On the other hand, *proper* time $\tau$ is always given without the factor $c$ (i.e. not as $c\tau$), that is in time units.
• Now there is no fundamental reason to use space units instead of time units, in particular for the $(-)$ case in which *timelike* intervals get positive squared metric. This would mean to specify points as
• $$(t,x/c,y/c,z/c) \tag{T}$$
• Now it seems clear to me why this option is not used: With the standard convention the 4-equivalents of 3-vectors get the same units as the corresponding 3-vectors. For example, the 4-momentum vector
• $$p=(mc \frac{\mathrm dt}{\mathrm d\tau}, • m\frac{\mathrm dx}{\mathrm d\tau}, • m\frac{\mathrm dy}{\mathrm d\tau}, • m\frac{\mathrm dz}{\mathrm d\tau})$$
• has actually the units of a momentum; in particular the space components directly give the 3-momentum.
• However I think there is a better way to do this. This is based on the observation that for coordinate systems like polar coordinates not all coordinates have the same unit, and the unit difference is then reflected in the metric. Therefore I see no reason not to do the same also for spacetime coordinates.
• In this scheme, the spacetime coordinates would be written simply as
• $$(t,x,y,z) \tag{M}$$
• and the corresponding metric (choosing the length to be in spatial units, like in the standard convention) would be
• $$g = \operatorname{diag}(-c^2,1,1,1) \tag{+'}$$
• or
• $$g = \operatorname{diag}(c^2,-1,-1,-1) \tag{-'}$$
• This convention would IMHO have several advantages:
• * Unlike the standard convention $(\mathrm S)$, $(\mathrm M)$ would be consistent with the fact that *proper* time is always given without the $c$ factor in the context of relativity.
• * But unlike $(\mathrm T)$ (and unlike the option to just use $c\tau$ for the proper time for consistency), the choice $(\mathrm M)$ keeps the usual units for the spacelike components of 4-vectors. For example, the four-momentum in that convention is
• $$p=(m\frac{\mathrm dt}{\mathrm d\tau}, • m\frac{\mathrm dx}{\mathrm d\tau}, • m\frac{\mathrm dy}{\mathrm d\tau}, • m\frac{\mathrm dz}{\mathrm d\tau})$$
• Clearly the space components are the 3-momentum in usual units. The time component (which gives the energy) isn't in conventional units (instead it is in mass units, thus giving the “relativistic mass” $E/c^2$), but it isn't in the standard convention either (where the time component is $E/c$).
• Now my question is: Is there any downside (other than “it's not what is commonly done”) to the convention $(\mathrm M)$?
#2: Post edited by celtschk‭ · 2021-01-10T12:20:43Z (over 1 year ago)
• In special relativity, spacetime coordinates are typically given as
• $$(ct, x, y, z) \tag{S}$$
• with the metric being either
• $$g = \operatorname{diag}(-1,1,1,1) \tag{+}$$
• or
• $$g = \operatorname{diag}(1,-1,-1,-1) \tag{-}$$
• depending on which sign convention the author prefers (the second one seems to be most commonly used one).
• The effect is that all coordinates, including the time coordinate, are given in space units. On the other hand, *proper* time $\tau$ is always given without the factor $c$ (i.e. not as $c\tau$), that is in time units.
• Now there is no fundamental reason to use space units instead of time units, in particular for the $(-)$ case in which *timelike* intervals get positive squared metric. This would mean to specify points as
• $$(t,x/c,y/c,z/c) \tag{T}$$
• Now it seems clear to me why this option is not used: With the standard convention the 4-equivalents of 3-vectors get the same units as the corresponding 3-vectors. For example, the 4-momentum vector
• $$p=(mc \frac{\mathrm dt}{\mathrm d\tau}, • m\frac{\mathrm dx}{\mathrm d\tau}, • m\frac{\mathrm dy}{\mathrm d\tau}, • m\frac{\mathrm dz}{\mathrm d\tau})$$
• has actually the units of a momentum; in particular the space components directly give the $3$-momentum.
• However I think there is a better way to do this. This is based on the observation that for coordinate systems like polar coordinates not all coordinates have the same unit, and the unit difference is then reflected in the metric. Therefore I see no reason not to do the same also for spacetime coordinates.
• In this scheme, the spacetime coordinates would be written simply as
• $$(t,x,y,z) \tag{M}$$
• and the corresponding metric (choosing the length to be in spatial units, like in the standard convention) would be
• $$g = \operatorname{diag}(-c^2,1,1,1) \tag{+'}$$
• or
• $$g = \operatorname{diag}(c^2,-1,-1,-1) \tag{-'}$$
• This convention would IMHO have several advantages:
• * Unlike the standard convention $(S)$, $(M)$ would be consistent with the fact that *proper* time is always given without the $c$ factor in the context of relativity.
• * But unlike $(T)$ (and unlike the option to just use $c\tau$ for the proper time for consistency), the choice $(M)$ keeps the usual units for the spacelike components of 4-vectors. For example, the four-momentum in that convention is
• $$p=(m\frac{\mathrm dt}{\mathrm d\tau}, • m\frac{\mathrm dx}{\mathrm d\tau}, • m\frac{\mathrm dy}{\mathrm d\tau}, • m\frac{\mathrm dz}{\mathrm d\tau})$$
• Clearly the space components are the 3-momentum in usual units. The time component (which gives the energy) isn't in conventional units (instead it is in mass units, thus giving the “relativistic mass” $E/c^2$), but it isn't in the standard convention either (where the time component is $E/c$).
• Now my question is: Is there any downside (other than “it's not what is commonly done”) to the convention $(M)$?
• In special relativity, spacetime coordinates are typically given as
• $$(ct, x, y, z) \tag{S}$$
• with the metric being either
• $$g = \operatorname{diag}(-1,1,1,1) \tag{+}$$
• or
• $$g = \operatorname{diag}(1,-1,-1,-1) \tag{-}$$
• depending on which sign convention the author prefers (the second one seems to be most commonly used one).
• The effect is that all coordinates, including the time coordinate, are given in space units. On the other hand, *proper* time $\tau$ is always given without the factor $c$ (i.e. not as $c\tau$), that is in time units.
• Now there is no fundamental reason to use space units instead of time units, in particular for the $(-)$ case in which *timelike* intervals get positive squared metric. This would mean to specify points as
• $$(t,x/c,y/c,z/c) \tag{T}$$
• Now it seems clear to me why this option is not used: With the standard convention the 4-equivalents of 3-vectors get the same units as the corresponding 3-vectors. For example, the 4-momentum vector
• $$p=(mc \frac{\mathrm dt}{\mathrm d\tau}, • m\frac{\mathrm dx}{\mathrm d\tau}, • m\frac{\mathrm dy}{\mathrm d\tau}, • m\frac{\mathrm dz}{\mathrm d\tau})$$
• has actually the units of a momentum; in particular the space components directly give the 3-momentum.
• However I think there is a better way to do this. This is based on the observation that for coordinate systems like polar coordinates not all coordinates have the same unit, and the unit difference is then reflected in the metric. Therefore I see no reason not to do the same also for spacetime coordinates.
• In this scheme, the spacetime coordinates would be written simply as
• $$(t,x,y,z) \tag{M}$$
• and the corresponding metric (choosing the length to be in spatial units, like in the standard convention) would be
• $$g = \operatorname{diag}(-c^2,1,1,1) \tag{+'}$$
• or
• $$g = \operatorname{diag}(c^2,-1,-1,-1) \tag{-'}$$
• This convention would IMHO have several advantages:
• * Unlike the standard convention $(S)$, $(M)$ would be consistent with the fact that *proper* time is always given without the $c$ factor in the context of relativity.
• * But unlike $(T)$ (and unlike the option to just use $c\tau$ for the proper time for consistency), the choice $(M)$ keeps the usual units for the spacelike components of 4-vectors. For example, the four-momentum in that convention is
• $$p=(m\frac{\mathrm dt}{\mathrm d\tau}, • m\frac{\mathrm dx}{\mathrm d\tau}, • m\frac{\mathrm dy}{\mathrm d\tau}, • m\frac{\mathrm dz}{\mathrm d\tau})$$
• Clearly the space components are the 3-momentum in usual units. The time component (which gives the energy) isn't in conventional units (instead it is in mass units, thus giving the “relativistic mass” $E/c^2$), but it isn't in the standard convention either (where the time component is $E/c$).
• Now my question is: Is there any downside (other than “it's not what is commonly done”) to the convention $(M)$?
#1: Initial revision by celtschk‭ · 2021-01-10T12:18:45Z (over 1 year ago)
Unusual way to write spacetime coordinates/metric: Is there any downside?
In special relativity, spacetime coordinates are typically given as
$$(ct, x, y, z) \tag{S}$$
with the metric being either
$$g = \operatorname{diag}(-1,1,1,1) \tag{+}$$
or
$$g = \operatorname{diag}(1,-1,-1,-1) \tag{-}$$
depending on which sign convention the author prefers (the second one seems to be most commonly used one).

The effect is that all coordinates, including the time coordinate, are given in space units. On the other hand, *proper* time $\tau$ is always given without the factor $c$ (i.e. not as $c\tau$), that is in time units.

Now there is no fundamental reason to use space units instead of time units, in particular for the $(-)$ case in which *timelike* intervals get positive squared metric. This would mean to specify points as
$$(t,x/c,y/c,z/c) \tag{T}$$

Now it seems clear to me why this option is not used: With the standard convention the 4-equivalents of 3-vectors get the same units as the corresponding 3-vectors. For example, the 4-momentum vector
$$p=(mc \frac{\mathrm dt}{\mathrm d\tau}, m\frac{\mathrm dx}{\mathrm d\tau}, m\frac{\mathrm dy}{\mathrm d\tau}, m\frac{\mathrm dz}{\mathrm d\tau})$$
has actually the units of a momentum; in particular the space components directly give the $3$-momentum.

However I think there is a better way to do this. This is based on the observation that for coordinate systems like polar coordinates not all coordinates have the same unit, and the unit difference is then reflected in the metric. Therefore I see no reason not to do the same also for spacetime coordinates.

In this scheme, the spacetime coordinates would be written simply as
$$(t,x,y,z) \tag{M}$$
and the corresponding metric (choosing the length to be in spatial units, like in the standard convention) would be
$$g = \operatorname{diag}(-c^2,1,1,1) \tag{+'}$$
or
$$g = \operatorname{diag}(c^2,-1,-1,-1) \tag{-'}$$

This convention would IMHO have several advantages:

* Unlike the standard convention $(S)$, $(M)$ would be consistent with the fact that *proper* time is always given without the $c$ factor in the context of relativity.

* But unlike $(T)$ (and unlike the option to just use $c\tau$ for the proper time for consistency), the choice $(M)$ keeps the usual units for the spacelike components of 4-vectors. For example, the four-momentum in that convention is
$$p=(m\frac{\mathrm dt}{\mathrm d\tau}, m\frac{\mathrm dx}{\mathrm d\tau}, m\frac{\mathrm dy}{\mathrm d\tau}, m\frac{\mathrm dz}{\mathrm d\tau})$$
Clearly the space components are the 3-momentum in usual units. The time component (which gives the energy) isn't in conventional units (instead it is in mass units, thus giving the “relativistic mass” $E/c^2$), but it isn't in the standard convention either (where the time component is $E/c$).

Now my question is: Is there any downside (other than “it's not what is commonly done”) to the convention $(M)$?